Quadratic Equations and the Quadratic Formula

41 questions

Question 1Question

What is the sum of the squares of the two real solutions to the quadratic equation 12x23x+2=0\frac{1}{2}x^2 - 3x + 2 = 0?

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Answer: 28

Answer

The sum of the squares of the two real solutions is 28.
Multiplying the equation 12x23x+2=0\frac{1}{2}x^2 - 3x + 2 = 0 by 2 yields x26x+4=0x^2 - 6x + 4 = 0. By Vieta's formulas, the sum of the solutions is x1+x2=6x_1 + x_2 = 6 and the product of the solutions is x1x2=4x_1 x_2 = 4. Using the algebraic identity x12+x22=(x1+x2)22x1x2x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2, the sum of the squares is 622(4)=368=286^2 - 2(4) = 36 - 8 = 28. Alternatively, using the quadratic formula on the simplified equation gives solutions of 3+53 + \sqrt{5} and 353 - \sqrt{5}. Squaring these values yields 14+6514 + 6\sqrt{5} and 146514 - 6\sqrt{5}, which sum to 28.

Step-by-Step Solution

1
Multiply the quadratic equation by 2 to clear the fraction.
x26x+4=0x^2 - 6x + 4 = 0
Simplifying the fractional coefficients makes the equation easier to analyze and solve.
2
Identify the sum and product of the roots using Vieta's formulas.
x1+x2=6x_1 + x_2 = 6 and x1x2=4x_1 x_2 = 4
For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is b/a-b/a and the product is c/ac/a.
3
Express the sum of the squares of the roots using the algebraic identity.
x12+x22=(x1+x2)22x1x2x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2
This identity allows the direct calculation of the sum of squares without needing to find the individual roots.
4
Substitute the values of the sum and product into the identity.
622(4)=368=286^2 - 2(4) = 36 - 8 = 28
Plugging in the sum of 6 and product of 4 yields the final numerical answer.

Key Concept

Vieta's formulas state that for a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is x1+x2=bax_1 + x_2 = -\frac{b}{a} and the product of the roots is x1x2=cax_1 x_2 = \frac{c}{a}. Symmetric functions of roots like x12+x22x_1^2 + x_2^2 can be expressed in terms of these values.

Alternative Method

Find the roots of the equation directly using the quadratic formula. After simplifying to x26x+4=0x^2 - 6x + 4 = 0, the roots are x=6±(6)24(1)(4)2=3±5x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(4)}}{2} = 3 \pm \sqrt{5}. Squaring both solutions gives (3+5)2=14+65(3 + \sqrt{5})^2 = 14 + 6\sqrt{5} and (35)2=1465(3 - \sqrt{5})^2 = 14 - 6\sqrt{5}. Adding these squares together yields (14+65)+(1465)=28(14 + 6\sqrt{5}) + (14 - 6\sqrt{5}) = 28.
Estimated Time:1m 30s
Question 2Question

A quadratic equation of the form ax2+bx+c=0a x^2 + b x + c = 0, where aa, bb, and cc are real constants and a>0a > 0, has a discriminant of 3737. If the sum of the roots of this equation is 5.55.5 and the product of the roots is 5.255.25, what is the value of the coefficient aa?

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Answer: 2

Answer

The value of the coefficient aa is 22.
By applying Vieta's formulas, we can write b=5.5ab = -5.5a and c=5.25ac = 5.25a. Plugging these into the discriminant formula gives D=(5.5a)24a(5.25a)=30.25a221a2=9.25a2D = (-5.5a)^2 - 4a(5.25a) = 30.25a^2 - 21a^2 = 9.25a^2. Setting the discriminant to 3737 results in 9.25a2=37    a2=49.25a^2 = 37 \implies a^2 = 4. Since the problem specifies a>0a > 0, taking the positive square root gives a=2a = 2.

Step-by-Step Solution

1
Express the coefficients bb and cc in terms of aa using Vieta's formulas.
b=5.5ab = -5.5a and c=5.25ac = 5.25a
The sum of the roots is ba-\frac{b}{a} and the product is ca\frac{c}{a}.
2
Substitute the expressions for bb and cc into the discriminant formula D=b24acD = b^2 - 4ac.
D=9.25a2D = 9.25a^2
Substituting the terms yields D=(5.5a)24a(5.25a)=30.25a221a2=9.25a2D = (-5.5a)^2 - 4a(5.25a) = 30.25a^2 - 21a^2 = 9.25a^2.
3
Equate the discriminant expression to 3737 and solve for aa.
a=2a = 2
Since D=37D = 37, we write 9.25a2=37    a2=49.25a^2 = 37 \implies a^2 = 4. Because aa must be positive, we find a=2a = 2.

Key Concept

Using the properties of quadratic roots (Vieta's formulas) and the definition of the discriminant to solve for coefficients.
Question 3Question

In the quadratic equation ax2+bx+c=0a x^2 + b x + c = 0, the coefficients aa, bb, and cc are real numbers, and a=23a = \frac{2}{3}. If one of the roots of this equation is x=12i2x = \frac{1}{2} - i\sqrt{2}, where i=1i = \sqrt{-1}, what is the value of the product bcbc?

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Answer: 1-1

Answer

1-1
For any quadratic equation with real coefficients, if a complex number is a root, its conjugate must also be a root. Thus, the two roots are x1=12i2x_1 = \frac{1}{2} - i\sqrt{2} and x2=12+i2x_2 = \frac{1}{2} + i\sqrt{2}. Using Vieta's formulas, the sum of the roots is x1+x2=1=bax_1 + x_2 = 1 = -\frac{b}{a}, which gives b=23b = -\frac{2}{3}. The product of the roots is x1x2=142i2=94=cax_1 x_2 = \frac{1}{4} - 2i^2 = \frac{9}{4} = \frac{c}{a}, which gives c=32c = \frac{3}{2}. The product bcbc is therefore (23)(32)=1(-\frac{2}{3})(\frac{3}{2}) = -1.

Step-by-Step Solution

1
Find the second root of the quadratic equation.
The second root is x2=12+i2x_2 = \frac{1}{2} + i\sqrt{2}.
Since the quadratic equation has real coefficients, complex roots must occur in conjugate pairs.
2
Calculate the sum of the roots to find the coefficient bb.
b=23b = -\frac{2}{3}
The sum of the roots is x1+x2=(12i2)+(12+i2)=1x_1 + x_2 = (\frac{1}{2} - i\sqrt{2}) + (\frac{1}{2} + i\sqrt{2}) = 1. By Vieta's formulas, x1+x2=bax_1 + x_2 = -\frac{b}{a}. Substituting a=23a = \frac{2}{3} gives 1=b2/3    b=231 = -\frac{b}{2/3} \implies b = -\frac{2}{3}.
3
Calculate the product of the roots to find the coefficient cc.
c=32c = \frac{3}{2}
The product of the roots is x1x2=(12i2)(12+i2)=(12)2(i2)2=142i2x_1 x_2 = (\frac{1}{2} - i\sqrt{2})(\frac{1}{2} + i\sqrt{2}) = (\frac{1}{2})^2 - (i\sqrt{2})^2 = \frac{1}{4} - 2i^2. Since i2=1i^2 = -1, the product is \frac{1}{4} - 2(-1) = \frac{9}{4}.ByVietasformulas,. By Vieta's formulas, x_1 x_2 = \frac{c}{a}.Substituting. Substituting a = \frac{2}{3}gives94=c2/3    c=32 gives \frac{9}{4} = \frac{c}{2/3} \implies c = \frac{3}{2}.
4
Calculate the product bcbc.
bc=1bc = -1
Multiplying the values of bb and cc gives bc=(23)(32)=1bc = (-\frac{2}{3})(\frac{3}{2}) = -1.

Key Concept

Relating the roots of a quadratic equation with real coefficients to its coefficients via Vieta's formulas and the Complex Conjugate Root Theorem.
Question 4Question

A projectile is launched from ground level with an initial upward velocity of vv meters per second. The height of the projectile in meters after tt seconds is modeled by the equation h(t)=92t2+vth(t) = -\frac{9}{2}t^2 + vt. A sensor is placed at a height of 3232 meters. What is the minimum integer value of vv for which the projectile is at or above the height of the sensor for a duration of at least 44 seconds?

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Answer: 30

Answer

The minimum integer value of the initial upward velocity vv is 30.
The height condition h(t)32h(t) \ge 32 leads to the quadratic inequality 9t22vt+6409t^2 - 2vt + 64 \le 0. The roots t1t_1 and t2t_2 of 9t22vt+64=09t^2 - 2vt + 64 = 0 represent the boundary times. The duration above 3232 meters is the difference of the roots: t2t1=b24aca=4v223049t_2 - t_1 = \frac{\sqrt{b^2 - 4ac}}{|a|} = \frac{\sqrt{4v^2 - 2304}}{9}. Setting this to be at least 44 seconds gives 4v2230494\frac{\sqrt{4v^2 - 2304}}{9} \ge 4. Squaring both sides after multiplying by 99 yields 4v2230412964v^2 - 2304 \ge 1296, which simplifies to 4v236004v^2 \ge 3600, or v2900v^2 \ge 900. Taking the positive square root gives v30v \ge 30, meaning the minimum integer value is 3030.

Step-by-Step Solution

1
Write the inequality representing the condition that the projectile is at or above the sensor height.
9t22vt+6409t^2 - 2vt + 64 \le 0
The height function h(t)=92t2+vth(t) = -\frac{9}{2}t^2 + vt must be greater than or equal to 3232 meters.
2
Use the formula for the difference of roots to represent the time duration.
t2t1=4v223049t_2 - t_1 = \frac{\sqrt{4v^2 - 2304}}{9}
The duration of time the projectile is above the sensor is the difference between the two times at which the projectile crosses the sensor height.
3
Set the duration to be at least 44 seconds and solve for vv.
v30v \ge 30
Solving the inequality 4v2230494\frac{\sqrt{4v^2 - 2304}}{9} \ge 4 yields the range of initial velocities.
4
Find the smallest integer in the solution set.
3030
Since the velocity must be an integer and at least 3030, the minimum possible integer value is 3030.

Key Concept

Solving quadratic inequalities and finding the difference of roots using the discriminant.
Question 5Question

A rectangular photograph has a length of xx inches and a width that is 1.51.5 inches shorter than its length. If the area of the photograph is 1010 square inches, what is the length, in inches, of the photograph?

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Answer: 4

Answer

4
The correct length is 4. When the length is 4 inches, the width is 1.5 inches shorter, which is 2.5 inches. The area is the product of these dimensions: 4 * 2.5 = 10 square inches, which matches the given condition.

Step-by-Step Solution

1
Set up the equation for the area of the photograph.
x(x1.5)=10x(x - 1.5) = 10
The area of a rectangle is found by multiplying its length (xx) by its width (x1.5x - 1.5).
2
Distribute the xx and subtract 1010 from both sides to write the equation in standard form.
x21.5x10=0x^2 - 1.5x - 10 = 0
Standard form for a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0.
3
Multiply the entire equation by 22 to clear the decimal coefficient.
2x23x20=02x^2 - 3x - 20 = 0
Working with integer coefficients makes factoring or applying the quadratic formula easier.
4
Factor the quadratic expression.
(2x+5)(x4)=0(2x + 5)(x - 4) = 0
Factoring helps find the roots of the equation directly.
5
Solve for xx by setting each factor to zero.
x=2.5x = -2.5 or x=4x = 4
If the product of two numbers is zero, at least one of the numbers must be zero.
6
Exclude the negative solution.
x=4x = 4
The length of a photograph must be a positive value.

Key Concept

Solving a quadratic equation by translating a word problem and factoring.
Question 6Question

For the quadratic equation 3x26x+2=03x^2 - 6x + 2 = 0, what is the value of the discriminant?

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Answer: 12

Answer

The discriminant of the quadratic equation is 12.
The discriminant of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 is determined by the expression b24acb^2 - 4ac. By substituting the coefficients a=3a = 3, b=6b = -6, and c=2c = 2 from the given equation 3x26x+2=03x^2 - 6x + 2 = 0, we calculate (6)24(3)(2)=3624=12(-6)^2 - 4(3)(2) = 36 - 24 = 12.

Step-by-Step Solution

1
Identify the coefficients of the quadratic equation
a=3a = 3, b=6b = -6, and c=2c = 2
The standard form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0.
2
Substitute the coefficients into the discriminant formula
D=(6)24(3)(2)D = (-6)^2 - 4(3)(2)
The discriminant formula is D=b24acD = b^2 - 4ac.
3
Simplify the expression to find the final value
3624=1236 - 24 = 12
Squaring 6-6 gives 3636 and multiplying 4×3×24 \times 3 \times 2 gives 2424. Subtracting 2424 from 3636 gives 1212.

Key Concept

Calculating the discriminant of a quadratic equation
Estimated Time:45s
Question 7Question

What are the solutions to the quadratic equation x252x+1=0x^2 - \frac{5}{2}x + 1 = 0?

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Answer: x=12x = \frac{1}{2} and x=2x = 2

Answer

The solutions are x=12x = \frac{1}{2} and x=2x = 2.
The correct answer provides the exact solutions to the equation. Multiplying the equation by 22 yields 2x25x+2=02x^2 - 5x + 2 = 0. Factoring this expression gives (2x1)(x2)=0(2x - 1)(x - 2) = 0, which yields the solutions x=12x = \frac{1}{2} and x=2x = 2.

Step-by-Step Solution

1
Multiply the entire equation by 22 to clear the fraction.
2x25x+2=02x^2 - 5x + 2 = 0
Converting the equation to integer coefficients simplifies factoring or applying the quadratic formula.
2
Identify the coefficients aa, bb, and cc to use in the quadratic formula.
a=2a = 2, b=5b = -5, c=2c = 2
These coefficients correspond to the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
3
Calculate the discriminant b24acb^2 - 4ac.
(5)24(2)(2)=2516=9(-5)^2 - 4(2)(2) = 25 - 16 = 9
The discriminant determines the number and type of solutions.
4
Substitute the coefficients and the discriminant into the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
x=(5)±92(2)=5±34x = \frac{-(-5) \pm \sqrt{9}}{2(2)} = \frac{5 \pm 3}{4}
This formula provides the exact solutions to any quadratic equation.
5
Evaluate the two cases for the plus/minus sign.
x=5+34=2x = \frac{5 + 3}{4} = 2 and x=534=12x = \frac{5 - 3}{4} = \frac{1}{2}
Splitting the formula into positive and negative branches yields the two distinct real roots.

Key Concept

Solving quadratic equations using the quadratic formula
Question 8Question

A right triangle has an area of 66 square units. The lengths of its legs, aa and bb, satisfy the quadratic equation a22.75ab+1.5b2=0a^2 - 2.75ab + 1.5b^2 = 0. If the length of the hypotenuse is an integer, what is the perimeter of the triangle?

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Answer: 12

Answer

The perimeter of the triangle is 12.
The correct answer is 12. By solving the quadratic relationship between the legs, we find two possible ratios: one where one leg is twice the other, and one where one leg is 0.75 times the other. Using the area of 6, the first case yields non-integer side lengths, while the second case yields legs of length 3 and 4. This results in an integer hypotenuse of 5, giving a perimeter of 3 + 4 + 5 = 12.

Step-by-Step Solution

1
Express the relation a22.75ab+1.5b2=0a^2 - 2.75ab + 1.5b^2 = 0 as a quadratic in terms of the ratio r=abr = \frac{a}{b}.
The equation becomes r22.75r+1.5=0r^2 - 2.75r + 1.5 = 0.
To find the relationship between the two legs of the triangle by solving for their ratio.
2
Apply the quadratic formula to solve for rr.
r=2r = 2 or r=0.75r = 0.75, meaning either a=2ba = 2b or a=0.75ba = 0.75b.
To determine the two possible linear relationships between the legs of the right triangle.
3
Substitute each ratio case into the area formula Area=12ab=6\text{Area} = \frac{1}{2}ab = 6, which simplifies to ab=12ab = 12.
For a=2ba = 2b, we get b=6b = \sqrt{6} and a=26a = 2\sqrt{6}. For a=0.75ba = 0.75b, we get b=4b = 4 and a=3a = 3.
To calculate the actual leg lengths for both geometric cases.
4
Calculate the hypotenuse c=a2+b2c = \sqrt{a^2 + b^2} for both cases to check which yields an integer value.
The first case yields c=30c = \sqrt{30}, which is not an integer. The second case yields c=32+42=5c = \sqrt{3^2 + 4^2} = 5, which is an integer.
To satisfy the constraint that the hypotenuse must be an integer, identifying the correct leg lengths as 33 and 44.
5
Calculate the perimeter of the triangle for the valid case.
The perimeter is 3+4+5=123 + 4 + 5 = 12.
To find the final requested value.

Key Concept

Solving quadratic relationships and applying the quadratic formula in geometric constraints.
Estimated Time:3m 0s
Question 9Question

What is the sum of the two solutions to the quadratic equation 2x27x4=02x^2 - 7x - 4 = 0?

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Answer: 3.5

Answer

The sum of the solutions is 3.53.5.
The sum of the solutions of the quadratic equation 2x27x4=02x^2 - 7x - 4 = 0 is 3.53.5. According to Vieta's formulas, the sum of the roots of a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0 is given by ba-\frac{b}{a}. Here, a=2a = 2 and b=7b = -7, so the sum is 72=3.5-\frac{-7}{2} = 3.5. Alternatively, solving the equation using the quadratic formula yields the roots 44 and 0.5-0.5, which sum to 3.53.5.

Step-by-Step Solution

1
Identify the coefficients from the quadratic equation 2x27x4=02x^2 - 7x - 4 = 0.
a=2a = 2, b=7b = -7, and c=4c = -4.
A quadratic equation in standard form is written as ax2+bx+c=0ax^2 + bx + c = 0.
2
Apply the sum of roots formula ba-\frac{b}{a}.
Sum =72=3.5= -\frac{-7}{2} = 3.5.
By Vieta's formulas, the sum of the roots of ax2+bx+c=0ax^2 + bx + c = 0 is ba-\frac{b}{a}.

Key Concept

Sum of roots of a quadratic equation using Vieta's formulas

Alternative Method

Solve the quadratic equation by factoring or using the quadratic formula. Factoring 2x27x4=02x^2 - 7x - 4 = 0 yields (2x+1)(x4)=0(2x + 1)(x - 4) = 0, which gives solutions x=0.5x = -0.5 and x=4x = 4. Adding these solutions together gives 0.5+4=3.5-0.5 + 4 = 3.5.
Estimated Time:1m 0s
Question 10Question

One of the solutions to the quadratic equation 0.5x2x+c=00.5x^2 - x + c = 0 is 1+i1 + i, where i=1i = \sqrt{-1} and cc is a constant. What is the value of cc?

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Answer: 1

Answer

The constant cc is equal to 11.
Substituting the given solution x=1+ix = 1 + i into the equation 0.5x2x+c=00.5x^2 - x + c = 0 gives 0.5(1+i)2(1+i)+c=00.5(1+i)^2 - (1+i) + c = 0. Since (1+i)2=1+2i+i2=1+2i1=2i(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i, the equation simplifies to 0.5(2i)1i+c=00.5(2i) - 1 - i + c = 0, which is i1i+c=0i - 1 - i + c = 0. Simplifying this yields 1+c=0-1 + c = 0, so the value of cc is 11.

Step-by-Step Solution

1
Substitute the solution x=1+ix = 1 + i into the quadratic equation 0.5x2x+c=00.5x^2 - x + c = 0.
0.5(1+i)2(1+i)+c=00.5(1+i)^2 - (1+i) + c = 0
Since 1+i1 + i is a solution to the equation, substituting it for xx must satisfy the equation.
2
Expand the squared term (1+i)2(1+i)^2 using the binomial square formula and the identity i2=1i^2 = -1.
(1+i)2=1+2i+i2=1+2i1=2i(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i
Expanding (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 with a=1a=1 and b=ib=i, and replacing i2i^2 with 1-1 simplifies the term.
3
Substitute 2i2i back into the equation and simplify.
0.5(2i)(1+i)+c=0i1i+c=01+c=00.5(2i) - (1+i) + c = 0 \Rightarrow i - 1 - i + c = 0 \Rightarrow -1 + c = 0
Multiplying 0.50.5 by 2i2i yields ii, distributing the negative sign across (1+i)(1+i) yields 1i-1 - i, and combining like terms leaves 1+c=0-1 + c = 0.
4
Solve the linear equation for cc.
c=1c = 1
Adding 11 to both sides isolates the variable cc.

Key Concept

Solving quadratic equations with complex roots by substitution and algebraic manipulation.
Question 11Question

If 32i3 - 2i is a root of the quadratic equation x2+bx+c=0x^2 + bx + c = 0, where bb and cc are real numbers and i=1i = \sqrt{-1}, what is the value of b+cb + c?

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Answer: 7

Answer

7
The correct answer is 77. Since the quadratic equation has real coefficients, the roots must be complex conjugates. The conjugate of the root 32i3 - 2i is 3+2i3 + 2i. Vieta's formulas show that the sum of the roots is b-b, meaning (32i)+(3+2i)=6=b(3 - 2i) + (3 + 2i) = 6 = -b, which yields b=6b = -6. The product of the roots is cc, meaning (32i)(3+2i)=94i2=94(1)=13(3 - 2i)(3 + 2i) = 9 - 4i^2 = 9 - 4(-1) = 13. Adding these coefficients together gives b+c=6+13=7b + c = -6 + 13 = 7.

Step-by-Step Solution

1
Identify the second root of the quadratic equation
The second root is 3+2i3 + 2i.
Since the quadratic equation has real coefficients, the Complex Conjugate Theorem dictates that if a complex number is a root, its complex conjugate must also be a root.
2
Determine the coefficient bb using the sum of the roots
b=6b = -6
According to Vieta's formulas, the sum of the roots of the equation x2+bx+c=0x^2 + bx + c = 0 is equal to b-b. Thus, (32i)+(3+2i)=6=b(3 - 2i) + (3 + 2i) = 6 = -b, which simplifies to b=6b = -6.
3
Determine the constant term cc using the product of the roots
c=13c = 13
According to Vieta's formulas, the product of the roots of the equation x2+bx+c=0x^2 + bx + c = 0 is equal to cc. Thus, (32i)(3+2i)=32(2i)2=94i2(3 - 2i)(3 + 2i) = 3^2 - (2i)^2 = 9 - 4i^2. Substituting i2=1i^2 = -1 yields 94(1)=9+4=139 - 4(-1) = 9 + 4 = 13.
4
Calculate the value of b+cb + c
b+c=7b + c = 7
Substitute the calculated values of bb and cc into the expression: 6+13=7-6 + 13 = 7.

Key Concept

The Complex Conjugate Theorem states that complex roots of polynomials with real coefficients occur in conjugate pairs. Vieta's formulas state that for a quadratic equation x2+bx+c=0x^2 + bx + c = 0, the sum of the roots is b-b and the product of the roots is cc.
Question 12Question

What is the value of the discriminant of the quadratic equation 3x2+5x2=03x^2 + 5x - 2 = 0?

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Answer: 49

Answer

The discriminant of the quadratic equation is 49.
The discriminant is calculated using the formula b24acb^2 - 4ac. For the equation 3x2+5x2=03x^2 + 5x - 2 = 0, the coefficients are a=3a = 3, b=5b = 5, and c=2c = -2. Substituting these yields 524(3)(2)=25(24)=25+24=495^2 - 4(3)(-2) = 25 - (-24) = 25 + 24 = 49.

Step-by-Step Solution

1
Identify the coefficients of the quadratic equation 3x2+5x2=03x^2 + 5x - 2 = 0 in the standard form ax2+bx+c=0ax^2 + bx + c = 0.
a=3a = 3, b=5b = 5, and c=2c = -2
To use the discriminant formula, we must first extract the constant coefficients corresponding to each term.
2
Substitute the coefficients into the discriminant formula D=b24acD = b^2 - 4ac.
D=524(3)(2)=25(24)=25+24=49D = 5^2 - 4(3)(-2) = 25 - (-24) = 25 + 24 = 49
Calculating the value of the discriminant provides the required solution.

Key Concept

Calculating the discriminant of a quadratic equation to determine the nature of its roots.
Question 13Question

For a constant kk, the quadratic equation 13x2k6x+(k10)=0\frac{1}{3}x^2 - \frac{k}{6}x + (k - 10) = 0 has two real roots, r1r_1 and r2r_2. If the sum of the reciprocals of the roots, 1r1+1r2\frac{1}{r_1} + \frac{1}{r_2}, is equal to 12-\frac{1}{2}, what is the value of kk?

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Answer: 7.5

Answer

The value of kk is 7.57.5.
By writing the sum of the reciprocals in terms of the sum and product of the roots, we find that 1r1+1r2=r1+r2r1r2\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}. Substituting the values from Vieta's formulas (r1+r2=k2r_1 + r_2 = \frac{k}{2} and r1r2=3(k10)r_1 r_2 = 3(k - 10)) yields k6(k10)=12\frac{k}{6(k - 10)} = -\frac{1}{2}. Solving this equation for kk results in k=7.5k = 7.5.

Step-by-Step Solution

1
Identify the quadratic coefficients in terms of the constant kk.
a=13a = \frac{1}{3}, b=k6b = -\frac{k}{6}, and c=k10c = k - 10
Applying Vieta's formulas requires the coefficients from the standard form ax2+bx+c=0ax^2 + bx + c = 0.
2
Determine the sum and the product of the roots using Vieta's formulas.
r1+r2=k2r_1 + r_2 = \frac{k}{2} and r1r2=3(k10)r_1 r_2 = 3(k - 10)
Vieta's formulas state that the sum of the roots is ba-\frac{b}{a} and the product of the roots is ca\frac{c}{a}.
3
Express the sum of the reciprocals of the roots in terms of kk.
1r1+1r2=r1+r2r1r2=k6(k10)\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2} = \frac{k}{6(k - 10)}
Finding a common denominator allows the sum of the reciprocals to be written as the ratio of the sum of the roots to the product of the roots.
4
Equate the expression to 12-\frac{1}{2} and solve the resulting equation for kk.
k=7.5k = 7.5
Solving k6(k10)=12\frac{k}{6(k - 10)} = -\frac{1}{2} gives 2k=6(k10)    8k=60    k=7.52k = -6(k - 10) \implies 8k = 60 \implies k = 7.5.

Key Concept

Vieta's Formulas and Algebraic Relationships of Roots
Estimated Time:2m 30s
Question 14Question

The length of a rectangle is 33 inches greater than its width. If the area of the rectangle is 1010 square inches, what is the width of the rectangle, in inches?

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Answer: 22

Answer

The width of the rectangle is 22 inches.
The correct answer is the option stating that the width is 22 inches. This is obtained by expressing the length as w+3w+3 and setting the area equation to w(w+3)=10w(w+3) = 10. Expanding and setting it to zero gives the quadratic equation w2+3w10=0w^2 + 3w - 10 = 0, which factors to (w+5)(w2)=0(w+5)(w-2) = 0. Since width must be positive, the only valid solution is 22.

Step-by-Step Solution

1
Define variables for the dimensions of the rectangle based on the problem statement.
Let the width of the rectangle be ww inches. The length of the rectangle is w+3w + 3 inches.
The problem states that the length is 33 inches greater than the width.
2
Set up an equation representing the area of the rectangle.
The area is the product of width and length: w(w+3)=10w(w + 3) = 10.
The area of a rectangle is calculated as width times length, and the given area is 1010 square inches.
3
Rewrite the equation in standard quadratic form aw2+bw+c=0aw^2 + bw + c = 0.
Distribute ww to get w2+3w=10w^2 + 3w = 10, then subtract 1010 from both sides: w2+3w10=0w^2 + 3w - 10 = 0.
Standard quadratic form is required to solve the equation by factoring or using the quadratic formula.
4
Factor the quadratic equation.
(w+5)(w2)=0(w + 5)(w - 2) = 0.
Finding factors that multiply to 10-10 and add to 33 allows us to find the roots.
5
Solve for ww and apply real-world constraints.
w=5w = -5 or w=2w = 2. Since a physical width must be positive, discard 5-5, leaving w=2w = 2.
A dimension of a geometric shape cannot be negative.

Key Concept

Solving quadratic equations derived from geometric word problems by factoring or using the quadratic formula.
Question 15Question

For a real number aa, the quadratic equation x2+ax+(a3)2=0x^2 + ax + (a - 3)^2 = 0 has complex roots. One of these roots is 2+ib2 + i\sqrt{b}, where bb is a positive real number and i=1i = \sqrt{-1}. What is the value of a+ba + b?

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Answer: 41

Answer

The value of a+ba + b is 4141.
The correct answer is 4141. Since the quadratic equation has real coefficients, its complex roots must occur in conjugate pairs. Thus, the roots are 2+ib2 + i\sqrt{b} and 2ib2 - i\sqrt{b}. Their sum is 44, which by Vieta's formulas equals a-a, giving a=4a = -4. The product of the roots is 4+b4 + b, which equals the constant term (a3)2=(43)2=49(a - 3)^2 = (-4 - 3)^2 = 49. Solving for bb gives b=45b = 45. Therefore, a+b=4+45=41a + b = -4 + 45 = 41.

Step-by-Step Solution

1
Identify the second root using the complex conjugate root theorem.
The conjugate root is 2ib2 - i\sqrt{b}.
Since the coefficients of the quadratic equation are real, complex roots must occur in conjugate pairs.
2
Find the value of aa using the sum of the roots.
a=4a = -4
By Vieta's formulas, the sum of the roots is equal to a-a. The sum of the conjugate roots is (2+ib)+(2ib)=4(2 + i\sqrt{b}) + (2 - i\sqrt{b}) = 4, so a=4-a = 4.
3
Calculate the constant term of the quadratic equation.
The constant term is 4949.
Substitute a=4a = -4 into the constant term (a3)2(a - 3)^2 to get (43)2=(7)2=49(-4 - 3)^2 = (-7)^2 = 49.
4
Find the value of bb using the product of the roots.
b=45b = 45
By Vieta's formulas, the product of the roots is the constant term. The product is (2+ib)(2ib)=4i2b=4+b(2 + i\sqrt{b})(2 - i\sqrt{b}) = 4 - i^2 b = 4 + b (since i2=1i^2 = -1). Equating this to the constant term gives 4+b=494 + b = 49, so b=45b = 45.
5
Calculate the final value of a+ba + b.
4141
Substitute the values of aa and bb: a+b=4+45=41a + b = -4 + 45 = 41.

Key Concept

Vieta's formulas and complex conjugate roots of quadratic equations with real coefficients.
Question 16Question

If the quadratic equation x2+kx+25=0x^2 + kx + 25 = 0 has exactly one real solution, and k>0k > 0, what is the value of kk?

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Answer: 10

Answer

The value of kk is 1010.
For any quadratic equation in standard form, ax2+bx+c=0ax^2 + bx + c = 0, the equation has exactly one real solution when the discriminant b24acb^2 - 4ac is equal to 00. Here, the equation is x2+kx+25=0x^2 + kx + 25 = 0, so a=1a = 1, b=kb = k, and c=25c = 25. Setting the discriminant to 00 gives k24(1)(25)=0k^2 - 4(1)(25) = 0, which simplifies to k2100=0k^2 - 100 = 0. Solving for kk yields k=±10k = \pm 10. Since we are given that k>0k > 0, the only valid solution is 1010.

Step-by-Step Solution

1
Set the discriminant of the quadratic equation to zero.
b24ac=0b^2 - 4ac = 0
A quadratic equation has exactly one real solution if and only if its discriminant is zero.
2
Identify coefficients aa, bb, and cc from the equation x2+kx+25=0x^2 + kx + 25 = 0 and substitute them into the discriminant equation.
k24(1)(25)=0k^2 - 4(1)(25) = 0
For x2+kx+25=0x^2 + kx + 25 = 0, the coefficients are a=1a=1, b=kb=k, and c=25c=25.
3
Simplify the equation and solve for kk.
k2=100k^2 = 100, so k=10k = 10 or k=10k = -10
Simplifying k2100=0k^2 - 100 = 0 gives k2=100k^2 = 100.
4
Apply the constraint k>0k > 0 to find the final value.
k=10k = 10
The problem specifies that kk must be greater than zero.

Key Concept

Using the discriminant (b24acb^2 - 4ac) of a quadratic equation to determine when there is exactly one real solution.
Estimated Time:45s
Question 17Question

For a real constant kk, the quadratic equation x22(k2)x+(k23k+2)=0x^2 - 2(k - 2)x + (k^2 - 3k + 2) = 0 has two real roots, r1r_1 and r2r_2. What is the value of kk that minimizes the sum of the squares of these roots, r12+r22r_1^2 + r_2^2?

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Answer: 2

Answer

The value of kk that minimizes the sum of the squares of the roots is 2.
Applying Vieta's formulas and algebraic identities, the sum of the squares of the roots is expressed as 2k210k+122k^2 - 10k + 12. The discriminant condition for the roots to be real requires k2k \leq 2. Since the vertex of the upward-opening parabola 2k210k+122k^2 - 10k + 12 is at k=2.5k = 2.5, the function is strictly decreasing for all k2k \leq 2. Thus, the minimum value on the interval k2k \leq 2 occurs at the boundary k=2k = 2.

Step-by-Step Solution

1
Use Vieta's formulas to find the sum and product of the roots.
r1+r2=2(k2)r_1 + r_2 = 2(k - 2) and r1r2=k23k+2r_1 r_2 = k^2 - 3k + 2
Vieta's formulas relate the roots of a quadratic equation Ax2+Bx+C=0Ax^2 + Bx + C = 0 to its coefficients by r1+r2=B/Ar_1 + r_2 = -B/A and r1r2=C/Ar_1 r_2 = C/A.
2
Express the sum of the squares of the roots, r12+r22r_1^2 + r_2^2, in terms of kk.
r12+r22=2k210k+12r_1^2 + r_2^2 = 2k^2 - 10k + 12
Since r12+r22=(r1+r2)22r1r2r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2, substituting the Vieta relations gives [2(k2)]22(k23k+2)=4(k24k+4)2k2+6k4=2k210k+12[2(k-2)]^2 - 2(k^2 - 3k + 2) = 4(k^2 - 4k + 4) - 2k^2 + 6k - 4 = 2k^2 - 10k + 12.
3
Determine the condition for the quadratic equation to have real roots using the discriminant.
Δ=4k+80    k2\Delta = -4k + 8 \geq 0 \implies k \leq 2
For the roots r1r_1 and r2r_2 to be real, the discriminant Δ=B24AC\Delta = B^2 - 4AC must be greater than or equal to 0.
4
Minimize the quadratic function f(k)=2k210k+12f(k) = 2k^2 - 10k + 12 subject to the constraint k2k \leq 2.
k=2k = 2
The parabola f(k)f(k) opens upwards with its vertex at k=2.5k = 2.5. For k2k \leq 2, the function is strictly decreasing, meaning its minimum value on this interval occurs at the upper boundary, k=2k = 2.

Key Concept

Quadratic Equations and the Quadratic Formula
Question 18Question

For a certain quadratic equation 0.25x2kx+4.5=00.25x^2 - kx + 4.5 = 0, where kk is a constant, the difference between the two real solutions is exactly 33. What is the positive value of kk?

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Answer: 2.25

Answer

The positive value of kk is 2.252.25.
By applying the root difference formula x1x2=b24aca|x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|} with a=0.25a = 0.25, b=kb = -k, and c=4.5c = 4.5, we get the equation k24.50.25=3\frac{\sqrt{k^2 - 4.5}}{0.25} = 3. This simplifies to 4k24.5=34\sqrt{k^2 - 4.5} = 3. Dividing by 44 and squaring both sides gives k24.5=0.5625k^2 - 4.5 = 0.5625, which means k2=5.0625k^2 = 5.0625. Taking the positive square root yields k=2.25k = 2.25.

Step-by-Step Solution

1
Identify the coefficients of the quadratic equation and express the formula for the difference of the roots.
The coefficients are a=0.25a = 0.25, b=kb = -k, and c=4.5c = 4.5. The difference between the roots is x1x2=b24aca|x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}.
This sets up the algebraic relationship between the root difference and the coefficients of the quadratic equation.
2
Substitute the coefficients and the given root difference of 33 into the difference formula.
(k)24(0.25)(4.5)0.25=3\frac{\sqrt{(-k)^2 - 4(0.25)(4.5)}}{0.25} = 3
This creates an equation in terms of the variable kk using the given problem constraints.
3
Simplify the equation and isolate the radical term.
4k24.5=3    k24.5=0.754\sqrt{k^2 - 4.5} = 3 \implies \sqrt{k^2 - 4.5} = 0.75
Simplifying the fraction makes it easier to isolate the radical prior to squaring.
4
Square both sides of the equation to eliminate the radical and solve for k2k^2.
k24.5=0.5625    k2=5.0625k^2 - 4.5 = 0.5625 \implies k^2 = 5.0625
Squaring is the inverse operation of the square root, allowing us to solve for k2k^2.
5
Calculate the positive square root of 5.06255.0625 to find the value of kk.
k=2.25k = 2.25
Since the question asks for the positive value of kk, we choose the positive square root.

Key Concept

Relating the difference of the roots of a quadratic equation to its coefficients using the discriminant and the quadratic formula.
Question 19Question

For what value of the constant kk does the quadratic equation 0.5x21.5x+k=00.5x^2 - 1.5x + k = 0 have two complex roots of the form a±2ia \pm 2i, where aa is a real number?

Show answer & explanation

Answer: 3.125

Answer

3.125
Applying the quadratic formula to 0.5x21.5x+k=00.5x^2 - 1.5x + k = 0 yields x=1.5±2.252kx = 1.5 \pm \sqrt{2.25 - 2k}. For the roots to be of the form a±2ia \pm 2i, the imaginary component must be 2i2i, which requires the radicand 2.252k2.25 - 2k to equal (2i)2=4(2i)^2 = -4. Solving the equation 2.252k=42.25 - 2k = -4 yields k=3.125k = 3.125. This corresponds to the option containing 3.1253.125.

Step-by-Step Solution

1
Identify the coefficients aa, bb, and cc of the given quadratic equation 0.5x21.5x+k=00.5x^2 - 1.5x + k = 0.
a=0.5a = 0.5, b=1.5b = -1.5, and c=kc = k.
These coefficients are required to substitute into the quadratic formula.
2
Apply the quadratic formula, x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, to express the roots of the equation.
x=(1.5)±(1.5)24(0.5)k2(0.5)=1.5±2.252k1=1.5±2.252kx = \frac{-(-1.5) \pm \sqrt{(-1.5)^2 - 4(0.5)k}}{2(0.5)} = \frac{1.5 \pm \sqrt{2.25 - 2k}}{1} = 1.5 \pm \sqrt{2.25 - 2k}.
Substituting the coefficients simplifies the expression for the roots of the quadratic equation.
3
Set the radical term equal to the imaginary component 2i2i to find the value of kk that yields complex roots of the form a±2ia \pm 2i.
2.252k=2i2.252k=(2i)22.252k=4\sqrt{2.25 - 2k} = 2i \Rightarrow 2.25 - 2k = (2i)^2 \Rightarrow 2.25 - 2k = -4.
For the roots to have an imaginary part of 2i2i, the term under the square root must equal (2i)2(2i)^2, which is 4-4.
4
Solve the linear equation 2.252k=42.25 - 2k = -4 for kk.
2k=42.252k=6.25k=3.125-2k = -4 - 2.25 \Rightarrow -2k = -6.25 \Rightarrow k = 3.125.
Isolating kk gives the constant value that satisfies the root condition.

Key Concept

Using the quadratic formula and discriminant to determine the value of a parameter for a specified type of complex roots.
Question 20Question

A model rocket is launched vertically upward from a platform. Its height, hh, in meters, tt seconds after launch is modeled by the equation h=4.9t2+v0t+2h = -4.9t^2 + v_0 t + 2, where v0v_0 is the constant initial upward velocity in meters per second. If the rocket never reaches a height of 1212 meters, which of the following inequalities represents all possible values of v0v_0?

Show answer & explanation

Answer: 0<v0<140 < v_0 < 14

Answer

0<v0<140 < v_0 < 14
The correct answer is the range of values where the velocity is between 0 and 14. Setting the rocket's height equal to 12 meters gives the equation 4.9t2+v0t+2=12-4.9t^2 + v_0 t + 2 = 12. Subtracting 12 from both sides results in 4.9t2+v0t10=0-4.9t^2 + v_0 t - 10 = 0. Since the rocket never reaches 12 meters, this quadratic equation has no real solutions, meaning its discriminant must be negative. Calculating the discriminant yields v024(4.9)(10)<0v_0^2 - 4(-4.9)(-10) < 0, which simplifies to v02196<0v_0^2 - 196 < 0. Solving for a positive velocity gives 0<v0<140 < v_0 < 14.

Step-by-Step Solution

1
Set up the equation for the rocket reaching a height of 12 meters.
4.9t2+v0t+2=12-4.9t^2 + v_0 t + 2 = 12
To find when the rocket reaches 12 meters, we set the height equation equal to 12.
2
Rearrange the equation into standard quadratic form.
4.9t2+v0t10=0-4.9t^2 + v_0 t - 10 = 0
Subtracting 12 from both sides of the equation sets it to 0, which is the standard form at2+bt+c=0at^2 + bt + c = 0.
3
Apply the condition for no real solutions using the discriminant.
Δ=v024(4.9)(10)<0\Delta = v_0^2 - 4(-4.9)(-10) < 0
For the rocket to never reach 12 meters, the quadratic equation must have no real solutions. This requires the discriminant Δ=b24ac\Delta = b^2 - 4ac to be strictly negative.
4
Solve the inequality for the positive velocity v0v_0.
v02196<0    v02<196    v0<14v_0^2 - 196 < 0 \implies v_0^2 < 196 \implies v_0 < 14
Multiplying the constants yields 4×4.9×10=1964 \times 4.9 \times 10 = 196. Taking the square root gives v0<14v_0 < 14. Since initial upward velocity must be positive, 0<v0<140 < v_0 < 14.

Key Concept

Using the discriminant of a quadratic equation to determine the number of real solutions in a physical context.
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