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Quadratic Equations and Polynomial Factoring

38 questions

All types38Multiple choice20Numeric18
Question 1Question →

If rr and ss are the two distinct real roots of the quadratic equation x2−6x+4=0x^2 - 6x + 4 = 0, what is the value of r3+s3r+s\frac{r^3 + s^3}{r + s}?

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Answer: 2424

Answer

The value of r3+s3r+s\frac{r^3 + s^3}{r + s} is 2424.
Using Vieta's formulas for x2−6x+4=0x^2 - 6x + 4 = 0, the sum of the roots is r+s=6r + s = 6 and the product is rs=4rs = 4. Factoring r3+s3r^3 + s^3 gives (r+s)(r2−rs+s2)(r + s)(r^2 - rs + s^2). Dividing by (r+s)(r + s) leaves r2−rs+s2r^2 - rs + s^2, which can be rewritten as (r+s)2−3rs(r + s)^2 - 3rs. Substituting the Vieta values yields 62−3(4)=36−12=246^2 - 3(4) = 36 - 12 = 24.

Step-by-Step Solution

1
Apply Vieta's formulas to the given quadratic equation x2−6x+4=0x^2 - 6x + 4 = 0.
The sum of the roots is r+s=6r + s = 6, and the product of the roots is rs=4rs = 4.
For a quadratic equation x2+bx+c=0x^2 + bx + c = 0, Vieta's formulas state that the sum of roots is −b-b and the product of roots is cc.
2
Factor the sum of cubes expression r3+s3r^3 + s^3.
r3+s3r+s=(r+s)(r2−rs+s2)r+s=r2−rs+s2\frac{r^3 + s^3}{r + s} = \frac{(r + s)(r^2 - rs + s^2)}{r + s} = r^2 - rs + s^2.
The sum of cubes factors algebraically into (r+s)(r2−rs+s2)(r + s)(r^2 - rs + s^2), and r+s=6≠0r + s = 6 \neq 0 allows cancellation.
3
Express r2−rs+s2r^2 - rs + s^2 in terms of (r+s)(r + s) and rsrs.
r2−rs+s2=(r+s)2−3rsr^2 - rs + s^2 = (r + s)^2 - 3rs.
Since (r+s)2=r2+2rs+s2(r + s)^2 = r^2 + 2rs + s^2, subtracting 3rs3rs yields r2−rs+s2r^2 - rs + s^2.
4
Substitute the known values r+s=6r + s = 6 and rs=4rs = 4 into the expression.
62−3(4)=36−12=246^2 - 3(4) = 36 - 12 = 24.
Evaluating the algebraic expression yields the final value.

Key Concept

Polynomial Factoring and Vieta's Formulas for Quadratic Equations
Estimated Time:1m 30s
Question 2Question →

What is the sum of all real solutions to the equation 2x2−5x+3=1−x\sqrt{2x^2 - 5x + 3} = 1 - x?

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Answer: 1

Answer

1
Squaring both sides of 2x2−5x+3=1−x\sqrt{2x^2 - 5x + 3} = 1 - x yields 2x2−5x+3=(1−x)2=x2−2x+12x^2 - 5x + 3 = (1 - x)^2 = x^2 - 2x + 1. Moving all terms to one side yields the quadratic equation x2−3x+2=0x^2 - 3x + 2 = 0, which factors as (x−1)(x−2)=0(x - 1)(x - 2) = 0. The algebraic solutions are x=1x = 1 and x=2x = 2. Testing x=1x = 1 in the original equation gives 0=0\sqrt{0} = 0, which is valid. Testing x=2x = 2 gives 1=−1\sqrt{1} = -1, which is invalid because the principal square root cannot be negative. Thus, x=1x = 1 is the sole valid real solution, and its sum is 1.

Step-by-Step Solution

1
Square both sides of the equation to eliminate the radical
2x2−5x+3=(1−x)2=1−2x+x22x^2 - 5x + 3 = (1 - x)^2 = 1 - 2x + x^2
Squaring both sides converts the radical equation into a standard algebraic polynomial equation.
2
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0
x2−3x+2=0x^2 - 3x + 2 = 0
Subtracting (x2−2x+1)(x^2 - 2x + 1) from both sides collects all terms on one side.
3
Factor the quadratic expression to find candidate roots
(x−1)(x−2)=0  ⟹  x=1 or x=2(x - 1)(x - 2) = 0 \implies x = 1 \text{ or } x = 2
Factoring determines the values of xx that satisfy the squared equation.
4
Substitute candidate roots back into the original equation to check for extraneous solutions
For x=1x = 1: 2(1)2−5(1)+3=0=0\sqrt{2(1)^2 - 5(1) + 3} = \sqrt{0} = 0, and 1−1=01 - 1 = 0 (Valid).
For x=2x = 2: 2(2)2−5(2)+3=1=1\sqrt{2(2)^2 - 5(2) + 3} = \sqrt{1} = 1, but 1−2=−11 - 2 = -1 (Invalid, since 1≠−11 \neq -1).
Squaring an equation can introduce extraneous roots where the principal square root would equal a negative quantity.
5
Sum the valid real solutions
The only valid solution is x=1x = 1, so the sum is 1.
Only valid roots that satisfy the original equation may be summed.

Key Concept

Quadratic Factoring and Extraneous Solutions in Radical Equations
Estimated Time:1m 30s
Question 3Question →

What is the sum of all real solutions to the equation (x−3)2=4(x−3)(x - 3)^2 = 4(x - 3)?

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Answer: 10

Answer

The sum of all real solutions to the equation is 10.
To find all solutions to (x−3)2=4(x−3)(x - 3)^2 = 4(x - 3), bring all terms to the left side: (x−3)2−4(x−3)=0(x - 3)^2 - 4(x - 3) = 0. Factoring out (x−3)(x - 3) yields (x−3)(x−7)=0(x - 3)(x - 7) = 0. This gives two distinct real roots, x=3x = 3 and x=7x = 7. Adding these together yields a sum of 10.

Step-by-Step Solution

1
Rearrange the given equation so that all terms are on one side.
(x−3)2−4(x−3)=0(x - 3)^2 - 4(x - 3) = 0
Moving all terms to one side allows for factoring without losing variable solutions.
2
Factor out the common algebraic factor (x−3)(x - 3).
(x−3) [(x−3)−4]=0  ⟹  (x−3)(x−7)=0(x - 3)\,[(x - 3) - 4] = 0 \implies (x - 3)(x - 7) = 0
Factoring avoids the mistake of dividing by a variable expression.
3
Set each factor equal to zero to find the real roots.
x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3 and x−7=0  ⟹  x=7x - 7 = 0 \implies x = 7
A product of factors equals zero if and only if at least one factor is zero.
4
Calculate the sum of the real solutions.
3+7=103 + 7 = 10
The question asks for the sum of all distinct real solutions.

Key Concept

Factoring Quadratic Expressions and Avoiding Variable Cancellation
Estimated Time:1m 0s
Question 4Question →

If mm and nn are non-zero real numbers such that mm and nn are the roots of the quadratic equation x2+mx+n=0x^2 + mx + n = 0, what is the value of m−nm - n?

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Answer: 3

Answer

3
The correct answer is 3. By Vieta's formulas, for x2+mx+n=0x^2 + mx + n = 0, the sum of roots m+n=−mm + n = -m gives 2m+n=02m + n = 0, and the product of roots m⋅n=nm \cdot n = n gives m=1m = 1 since n≠0n \neq 0. Substituting m=1m = 1 into 2m+n=02m + n = 0 yields n=−2n = -2. Thus, m−n=1−(−2)=3m - n = 1 - (-2) = 3.

Step-by-Step Solution

1
Apply Vieta's formulas to express the sum and product of roots in terms of coefficients.
For the quadratic equation x2+mx+n=0x^2 + mx + n = 0, the sum of roots is m+n=−mm + n = -m, and the product of roots is m⋅n=nm \cdot n = n.
By Vieta's relations for a standard quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of roots is −ba-\frac{b}{a} and the product is ca\frac{c}{a}.
2
Solve the product relation for mm.
Since n≠0n \neq 0, divide both sides of m⋅n=nm \cdot n = n by nn to get m=1m = 1.
Because nn is specified to be non-zero, division by nn is valid and yields a unique non-zero value for mm.
3
Substitute m=1m = 1 into the sum relation to determine nn.
1+n=−1  ⟹  n=−21 + n = -1 \implies n = -2.
Rearranging m+n=−mm + n = -m gives 2m+n=02m + n = 0, so n=−2(1)=−2n = -2(1) = -2.
4
Calculate the target value m−nm - n.
m−n=1−(−2)=3m - n = 1 - (-2) = 3.
Subtracting −2-2 from 11 produces 1+2=31 + 2 = 3.

Key Concept

Vieta's Formulas and Root-Coefficient Relationships in Quadratic Equations
Estimated Time:2m 0s
Question 5Question →
What is the product of all real solutions to the equation (x2−6x+10)2−2(x2−6x+10)−35=0?(x^2 - 6x + 10)^2 - 2(x^2 - 6x + 10) - 35 = 0?
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Answer: 3

Answer

The product of all real solutions to the equation is 3.
Substituting u=x2−6x+10u = x^2 - 6x + 10 converts the given equation into u2−2u−35=0u^2 - 2u - 35 = 0, which factors as (u−7)(u+5)=0(u - 7)(u + 5) = 0. Substituting back yields two quadratic equations: x2−6x+3=0x^2 - 6x + 3 = 0 and x2−6x+15=0x^2 - 6x + 15 = 0. Checking the discriminants reveals that x2−6x+3=0x^2 - 6x + 3 = 0 has D=24>0D = 24 > 0, producing two real roots with product ca=3\frac{c}{a} = 3, whereas x2−6x+15=0x^2 - 6x + 15 = 0 has D=−24<0D = -24 < 0, producing no real roots. Therefore, the product of all real solutions is 3.

Step-by-Step Solution

1
Substitute a dummy variable to simplify the equation structure.
Let u=x2−6x+10u = x^2 - 6x + 10. The equation becomes u2−2u−35=0u^2 - 2u - 35 = 0.
Recognizing the repeated quadratic expression allows transforming a fourth-degree equation into a standard quadratic form.
2
Factor the quadratic equation in terms of uu.
(u−7)(u+5)=0  ⟹  u=7 or u=−5(u - 7)(u + 5) = 0 \implies u = 7 \text{ or } u = -5.
Finding the values of uu establishes the intermediate equations for xx.
3
Analyze the first case u=7u = 7 for real solutions.
x2−6x+3=0x^2 - 6x + 3 = 0. Discriminant D1=36−12=24>0D_1 = 36 - 12 = 24 > 0. Product of real roots is 31=3\frac{3}{1} = 3.
A positive discriminant guarantees two real roots, and Vieta's formulas give their product directly.
4
Analyze the second case u=−5u = -5 for real solutions.
x2−6x+15=0x^2 - 6x + 15 = 0. Discriminant D2=36−60=−24<0D_2 = 36 - 60 = -24 < 0. No real roots.
A negative discriminant indicates complex conjugate roots, which must be excluded when finding the product of real solutions.
5
Combine results to find the final product of all real solutions.
Product = 3.
Only the two roots from the first case are real, so their product is the product of all real solutions.

Key Concept

Solving disguised quadratics via algebraic substitution and testing discriminants to filter out non-real roots before applying Vieta's formulas.
Question 6Question →

If xx is a real number such that x2−5x+1=0x^2 - 5x + 1 = 0, what is the value of x2+1x2x^2 + \frac{1}{x^2}?

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Answer: 23

Answer

The value of x2+1x2x^2 + \frac{1}{x^2} is 23.
Dividing the quadratic equation x2−5x+1=0x^2 - 5x + 1 = 0 by xx yields x+1x=5x + \frac{1}{x} = 5. Squaring both sides of this identity gives (x+1x)2=x2+2+1x2=25\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} = 25. Subtracting 2 from both sides results in x2+1x2=23x^2 + \frac{1}{x^2} = 23.

Step-by-Step Solution

1
Divide the given quadratic equation x2−5x+1=0x^2 - 5x + 1 = 0 by xx
x−5+1x=0  ⟹  x+1x=5x - 5 + \frac{1}{x} = 0 \implies x + \frac{1}{x} = 5
Since x≠0x \neq 0, dividing by xx isolates the sum of xx and its reciprocal.
2
Square both sides of the expression x+1x=5x + \frac{1}{x} = 5
(x+1x)2=25  ⟹  x2+2+1x2=25\left(x + \frac{1}{x}\right)^2 = 25 \implies x^2 + 2 + \frac{1}{x^2} = 25
Expanding the binomial square produces the required quadratic sum along with a constant cross-term.
3
Isolate x2+1x2x^2 + \frac{1}{x^2} by subtracting 2 from 25
x2+1x2=23x^2 + \frac{1}{x^2} = 23
Subtracting the constant cross-term yields the exact requested numeric value.

Key Concept

Algebraic transformation of quadratic equations into reciprocal power sums
Question 7Question →

If x(x2−7x+12)=2(x2−7x+12)x(x^2 - 7x + 12) = 2(x^2 - 7x + 12), what is the sum of all distinct real values of xx that satisfy the equation?

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Answer: 9

Answer

9
To find all solutions to x(x2−7x+12)=2(x2−7x+12)x(x^2 - 7x + 12) = 2(x^2 - 7x + 12), subtract the right side from the left to obtain (x−2)(x2−7x+12)=0(x - 2)(x^2 - 7x + 12) = 0. Factoring the quadratic expression gives (x−2)(x−3)(x−4)=0(x - 2)(x - 3)(x - 4) = 0. Setting each factor to zero reveals three distinct real roots: 2, 3, and 4. The sum of these roots is 2+3+4=92 + 3 + 4 = 9.

Step-by-Step Solution

1
Rearrange the equation to set one side to zero
x(x2−7x+12)−2(x2−7x+12)=0x(x^2 - 7x + 12) - 2(x^2 - 7x + 12) = 0
Setting the equation to zero allows the application of the zero-product property without illegally dividing by variable expressions.
2
Factor out the common quadratic expression (x2−7x+12)(x^2 - 7x + 12)
(x−2)(x2−7x+12)=0(x - 2)(x^2 - 7x + 12) = 0
Both terms share the common factor (x2−7x+12)(x^2 - 7x + 12).
3
Factor the quadratic component (x2−7x+12)(x^2 - 7x + 12) into linear factors
(x−2)(x−3)(x−4)=0(x - 2)(x - 3)(x - 4) = 0
The quadratic factors into (x−3)(x−4)(x - 3)(x - 4) because (−3)×(−4)=12(-3) \times (-4) = 12 and (−3)+(−4)=−7(-3) + (-4) = -7.
4
Apply the zero-product property to find all distinct real roots and compute their sum
Roots are x=2x = 2, x=3x = 3, and x=4x = 4. Sum = 2+3+4=92 + 3 + 4 = 9.
Setting each linear factor to zero gives the solutions. Summing these solutions yields 9.

Key Concept

Polynomial Factoring and the Zero-Product Property
Estimated Time:2m 0s
Question 8Question →

What is the sum of all distinct real values of xx that satisfy the equation x(x2−5x+6)=2(x2−5x+6)x(x^2 - 5x + 6) = 2(x^2 - 5x + 6)?

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Answer: 5

Answer

The sum of all distinct real solutions is 5.
To solve the equation x(x2−5x+6)=2(x2−5x+6)x(x^2 - 5x + 6) = 2(x^2 - 5x + 6), move all terms to the left side to get x(x2−5x+6)−2(x2−5x+6)=0x(x^2 - 5x + 6) - 2(x^2 - 5x + 6) = 0. Factoring out (x2−5x+6)(x^2 - 5x + 6) gives (x−2)(x2−5x+6)=0(x - 2)(x^2 - 5x + 6) = 0. Factoring the quadratic part yields (x−2)(x−2)(x−3)=0(x - 2)(x - 2)(x - 3) = 0, or (x−2)2(x−3)=0(x - 2)^2(x - 3) = 0. The real roots are x=2x = 2 and x=3x = 3. Taking the sum of these distinct real roots gives 2+3=52 + 3 = 5.

Step-by-Step Solution

1
Move all terms to one side of the equation to avoid dividing by a variable expression.
x(x2−5x+6)−2(x2−5x+6)=0x(x^2 - 5x + 6) - 2(x^2 - 5x + 6) = 0
Dividing by (x2−5x+6)(x^2 - 5x + 6) directly would lose valid roots where x2−5x+6=0x^2 - 5x + 6 = 0.
2
Factor out the common quadratic expression (x2−5x+6)(x^2 - 5x + 6).
(x−2)(x2−5x+6)=0(x - 2)(x^2 - 5x + 6) = 0
Applying the distributive property isolates the linear factor (x−2)(x - 2).
3
Factor the quadratic term x2−5x+6x^2 - 5x + 6 into linear factors.
(x−2)(x−2)(x−3)=0  ⟹  (x−2)2(x−3)=0(x - 2)(x - 2)(x - 3) = 0 \implies (x - 2)^2(x - 3) = 0
The quadratic factors into (x−2)(x−3)(x - 2)(x - 3) because (−2)+(−3)=−5(-2) + (-3) = -5 and (−2)×(−3)=6(-2) \times (-3) = 6.
4
Identify all distinct real solutions and compute their sum.
Real solutions are x=2x = 2 and x=3x = 3. Distinct sum: 2+3=52 + 3 = 5.
The root x=2x = 2 has multiplicity 2, but only distinct values must be summed.

Key Concept

Polynomial Factoring and Variable Cancellation Rules
Estimated Time:1m 30s
Question 9Question →

If xx is a real number such that (x2−5x+5)x2−9x+20=1(x^2 - 5x + 5)^{x^2 - 9x + 20} = 1, what is the sum of all distinct real values of xx that satisfy the equation?

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Answer: 15

Answer

The sum of all distinct real values of xx that satisfy the equation is 15.
The equation (x2−5x+5)x2−9x+20=1(x^2 - 5x + 5)^{x^2 - 9x + 20} = 1 holds under three distinct conditions: when the exponent is 0 and the base is non-zero (x=4,5x = 4, 5), when the base is 1 (x=1,4x = 1, 4), and when the base is −1-1 provided the exponent is an even integer (x=2,3x = 2, 3, yielding even exponents 6 and 2 respectively). The set of distinct solutions is {1,2,3,4,5}\{1, 2, 3, 4, 5\}, and their sum is 15.

Step-by-Step Solution

1
Analyze the general algebraic conditions under which an expression of the form uv=1u^v = 1 holds for real numbers uu and vv.
The equation uv=1u^v = 1 is satisfied in three distinct cases: (1) v=0v = 0 and u≠0u \neq 0, (2) u=1u = 1, or (3) u=−1u = -1 with vv being an even integer.
Any non-zero real number raised to the power of 0 equals 1, 1 raised to any power equals 1, and −1-1 raised to an even integer power equals 1.
2
Evaluate Case 1 where the exponent x2−9x+20=0x^2 - 9x + 20 = 0.
Factoring gives (x−4)(x−5)=0(x - 4)(x - 5) = 0, yielding x=4x = 4 and x=5x = 5. Testing the base x2−5x+5x^2 - 5x + 5 at x=4x = 4 gives 1≠01 \neq 0, and at x=5x = 5 gives 5≠05 \neq 0. Thus x=4x = 4 and x=5x = 5 are valid solutions.
Exponent equal to zero produces u0=1u^0 = 1 as long as the base u≠0u \neq 0.
3
Evaluate Case 2 where the base x2−5x+5=1x^2 - 5x + 5 = 1.
Subtracting 1 gives x2−5x+4=0x^2 - 5x + 4 = 0, which factors as (x−1)(x−4)=0(x - 1)(x - 4) = 0, yielding x=1x = 1 and x=4x = 4. Both are valid.
Base equal to 1 raised to any real exponent equals 1.
4
Evaluate Case 3 where the base x2−5x+5=−1x^2 - 5x + 5 = -1 and check if the exponent is an even integer.
Adding 1 gives x2−5x+6=0x^2 - 5x + 6 = 0, which factors as (x−2)(x−3)=0(x - 2)(x - 3) = 0, yielding x=2x = 2 and x=3x = 3. Testing x=2x = 2 in the exponent gives 22−9(2)+20=62^2 - 9(2) + 20 = 6 (even). Testing x=3x = 3 gives 32−9(3)+20=23^2 - 9(3) + 20 = 2 (even). Thus x=2x = 2 and x=3x = 3 are also valid solutions.
Base −1-1 raised to an even integer power equals 1.
5
Combine all unique real values of xx and compute their sum.
The set of distinct real solutions is {1,2,3,4,5}\{1, 2, 3, 4, 5\}. Their sum is 1+2+3+4+5=151 + 2 + 3 + 4 + 5 = 15.
Summing all identified distinct valid roots.

Key Concept

Solving polynomial exponential equations of the form f(x)g(x)=1f(x)^{g(x)} = 1 by systematically testing base-exponent cases.
Estimated Time:2m 0s
Question 10Question →

If xx is a real number such that x2−5x+3=0x^2 - 5x + 3 = 0, what is the value of x2+9x2x^2 + \frac{9}{x^2}?

Show answer & explanation

Answer: 19

Answer

The value of x2+9x2x^2 + \frac{9}{x^2} is 19.
Dividing x2−5x+3=0x^2 - 5x + 3 = 0 by xx gives x+3x=5x + \frac{3}{x} = 5. Squaring both sides yields (x+3x)2=x2+2(x)(3x)+9x2=25\left(x + \frac{3}{x}\right)^2 = x^2 + 2(x)\left(\frac{3}{x}\right) + \frac{9}{x^2} = 25, which simplifies to x2+6+9x2=25x^2 + 6 + \frac{9}{x^2} = 25. Subtracting 6 from both sides yields the final answer of 19.

Step-by-Step Solution

1
Divide the quadratic equation by xx
x+3x=5x + \frac{3}{x} = 5
Since x=0x = 0 is not a root of x2−5x+3=0x^2 - 5x + 3 = 0, dividing by xx rearranges the terms into xx and its reciprocal component.
2
Square both sides of x+3x=5x + \frac{3}{x} = 5
x2+6+9x2=25x^2 + 6 + \frac{9}{x^2} = 25
Applying the algebraic identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 generates the required expression x2+9x2x^2 + \frac{9}{x^2} alongside a constant middle term.
3
Isolate x2+9x2x^2 + \frac{9}{x^2}
x2+9x2=19x^2 + \frac{9}{x^2} = 19
Subtracting the constant cross-term 6 from 25 yields the target value.

Key Concept

Algebraic manipulation of quadratic equations using reciprocal square identities
Estimated Time:2m 0s
Question 11Question →
What is the sum of all real solutions to the polynomial equation (x2−3x+1)2−3(x2−3x+1)+1=x?(x^2 - 3x + 1)^2 - 3(x^2 - 3x + 1) + 1 = x ?
Show answer & explanation

Answer: 6

Answer

The sum of all real solutions to the equation is 6.
The equation (x2−3x+1)2−3(x2−3x+1)+1=x(x^2 - 3x + 1)^2 - 3(x^2 - 3x + 1) + 1 = x represents f(f(x))=xf(f(x)) = x for f(x)=x2−3x+1f(x) = x^2 - 3x + 1. Expanding and setting to zero yields x4−6x3+8x2+2x−1=0x^4 - 6x^3 + 8x^2 + 2x - 1 = 0. Since solutions to f(x)=xf(x) = x (i.e., x2−4x+1=0x^2 - 4x + 1 = 0) also satisfy f(f(x))=xf(f(x)) = x, we can factor out x2−4x+1x^2 - 4x + 1 to get (x2−4x+1)(x2−2x−1)=0(x^2 - 4x + 1)(x^2 - 2x - 1) = 0. Both quadratic factors have positive discriminants (1212 and 88), ensuring four distinct real roots. By Vieta's formulas, the sum of roots from the first quadratic is 4 and from the second is 2, giving a total sum of 6.

Step-by-Step Solution

1
Express the equation as a polynomial in standard form
Expanding (x2−3x+1)2−3(x2−3x+1)+1−x=0(x^2 - 3x + 1)^2 - 3(x^2 - 3x + 1) + 1 - x = 0 yields x4−6x3+8x2+2x−1=0x^4 - 6x^3 + 8x^2 + 2x - 1 = 0.
Expanding allows the composite expression to be treated as a degree-4 polynomial.
2
Find a quadratic factor using fixed points of the iteration
Let f(x)=x2−3x+1f(x) = x^2 - 3x + 1. The equation has the form f(f(x))=xf(f(x)) = x. Any solution to f(x)=xf(x) = x must satisfy this equation. Solving x2−3x+1=xx^2 - 3x + 1 = x gives x2−4x+1=0x^2 - 4x + 1 = 0.
If f(x)=xf(x) = x, then f(f(x))=f(x)=xf(f(x)) = f(x) = x, so x2−4x+1x^2 - 4x + 1 must be a factor of the 4th-degree polynomial.
3
Factor the degree-4 polynomial completely
Polynomial division of x4−6x3+8x2+2x−1x^4 - 6x^3 + 8x^2 + 2x - 1 by x2−4x+1x^2 - 4x + 1 yields (x2−4x+1)(x2−2x−1)=0(x^2 - 4x + 1)(x^2 - 2x - 1) = 0.
Factoring reduces the 4th-degree equation into two quadratic equations.
4
Verify real roots and calculate their sum
For x2−4x+1=0x^2 - 4x + 1 = 0, discriminant D1=12>0D_1 = 12 > 0 and sum of roots is 4. For x2−2x−1=0x^2 - 2x - 1 = 0, discriminant D2=8>0D_2 = 8 > 0 and sum of roots is 2. The four roots are all real and distinct, giving a total sum of 4+2=64 + 2 = 6.
Vieta's formulas provide the sum of roots for each quadratic factor without needing to compute individual radical expressions.

Key Concept

Polynomial Factoring and Composite Quadratic Equations
Estimated Time:2m 30s
Question 12Question →

If xx is a real number satisfying the equation x2+18x2+3=8x^2 + \frac{18}{x^2 + 3} = 8, what is the value of x2x^2?

Show answer & explanation

Answer: 6

Answer

The value of x2x^2 is 6.
Substituting u=x2u = x^2 converts the equation to u+18u+3=8u + \frac{18}{u+3} = 8. Multiplying through by (u+3)(u+3) yields u2+3u+18=8u+24u^2 + 3u + 18 = 8u + 24, which simplifies to the quadratic equation u2−5u−6=0u^2 - 5u - 6 = 0. Factoring gives (u−6)(u+1)=0(u-6)(u+1) = 0, so u=6u = 6 or u=−1u = -1. Because xx must be a real number, x2x^2 must be non-negative, eliminating u=−1u = -1. Thus, x2=6x^2 = 6.

Step-by-Step Solution

1
Substitute a temporary variable u=x2u = x^2 into the given equation
u+18u+3=8u + \frac{18}{u + 3} = 8
Using substitution simplifies the fourth-degree rational equation into a standard quadratic structure.
2
Clear the denominator by multiplying the entire equation by (u+3)(u + 3)
u(u+3)+18=8(u+3)u(u + 3) + 18 = 8(u + 3), which expands to u2+3u+18=8u+24u^2 + 3u + 18 = 8u + 24
Eliminating fractions allows rearranging terms into polynomial form.
3
Rearrange terms into standard quadratic form au2+bu+c=0au^2 + bu + c = 0
u2−5u−6=0u^2 - 5u - 6 = 0
Moving all terms to one side prepares the expression for factoring.
4
Factor the quadratic polynomial into linear binomial factors
(u−6)(u+1)=0(u - 6)(u + 1) = 0, giving u=6u = 6 or u=−1u = -1
The roots of the transformed equation represent candidate values for x2x^2.
5
Apply the constraint that xx is a real number to test candidate values
Since x2≥0x^2 \geq 0 for all real xx, u=−1u = -1 is rejected, leaving u=x2=6u = x^2 = 6
Squaring any real number yields a non-negative result.

Key Concept

Solving rational equations reducible to quadratics via algebraic substitution and rejecting non-real roots.
Estimated Time:2m 0s
Question 13Question →

What is the sum of all distinct real solutions to the equation (x2−3x)2−2(x2−3x)−8=0(x^2 - 3x)^2 - 2(x^2 - 3x) - 8 = 0?

Show answer & explanation

Answer: 6

Answer

The sum of all distinct real solutions is 6.
By defining u=x2−3xu = x^2 - 3x, the original equation reduces to u2−2u−8=0u^2 - 2u - 8 = 0, which factors as (u−4)(u+2)=0(u - 4)(u + 2) = 0, giving u=4u = 4 or u=−2u = -2. Substituting back for xx gives two quadratic equations: x2−3x−4=0x^2 - 3x - 4 = 0 (yielding roots x=4x = 4 and x=−1x = -1) and x2−3x+2=0x^2 - 3x + 2 = 0 (yielding roots x=1x = 1 and x=2x = 2). The four distinct real roots are −1,1,2,-1, 1, 2, and 44, and their sum is −1+1+2+4=6-1 + 1 + 2 + 4 = 6.

Step-by-Step Solution

1
Use algebraic substitution to simplify the disguised quadratic equation.
Let u=x2−3xu = x^2 - 3x. The equation becomes u2−2u−8=0u^2 - 2u - 8 = 0.
Recognizing the repeating quadratic block x2−3xx^2 - 3x transforms the 4th-degree polynomial into a standard 2nd-degree quadratic in terms of uu.
2
Factor the quadratic equation in terms of uu to find its roots.
(u−4)(u+2)=0  ⟹  u=4 or u=−2(u - 4)(u + 2) = 0 \implies u = 4 \text{ or } u = -2.
Finding the values of uu establishes the two quadratic equations in terms of xx that need to be solved.
3
Substitute x2−3xx^2 - 3x back for uu and solve for xx in Case 1 (u=4u = 4).
x2−3x=4  ⟹  x2−3x−4=0  ⟹  (x−4)(x+1)=0  ⟹  x=4 or x=−1x^2 - 3x = 4 \implies x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0 \implies x = 4 \text{ or } x = -1.
The discriminant b2−4ac=(−3)2−4(1)(−4)=25>0b^2 - 4ac = (-3)^2 - 4(1)(-4) = 25 > 0, so there are two distinct real solutions: x=4x = 4 and x=−1x = -1.
4
Substitute x2−3xx^2 - 3x back for uu and solve for xx in Case 2 (u=−2u = -2).
x2−3x=−2  ⟹  x2−3x+2=0  ⟹  (x−1)(x−2)=0  ⟹  x=1 or x=2x^2 - 3x = -2 \implies x^2 - 3x + 2 = 0 \implies (x - 1)(x - 2) = 0 \implies x = 1 \text{ or } x = 2.
The discriminant b2−4ac=(−3)2−4(1)(2)=1>0b^2 - 4ac = (-3)^2 - 4(1)(2) = 1 > 0, so there are two distinct real solutions: x=1x = 1 and x=2x = 2.
5
Sum all distinct real solutions.
−1+1+2+4=6-1 + 1 + 2 + 4 = 6.
All four solutions {−1,1,2,4}\{-1, 1, 2, 4\} are distinct and real.

Key Concept

Solving Disguised Quadratic Equations via Algebraic Substitution
Estimated Time:1m 30s
Question 14Question →

If the quadratic equation 3x2−kx+48=03x^2 - kx + 48 = 0 has equal real roots, where k>0k > 0, and the roots of the quadratic equation x2+kx−m=0x^2 + kx - m = 0 differ by 10, what is the value of mm?

Show answer & explanation

Answer: -119

Answer

The value of mm is −119-119.
Setting the discriminant of 3x2−kx+48=03x^2 - kx + 48 = 0 equal to zero gives k2−4(3)(48)=0k^2 - 4(3)(48) = 0, yielding k=24k = 24 since k>0k > 0. Substituting k=24k = 24 into the second equation gives x2+24x−m=0x^2 + 24x - m = 0. According to Vieta's formulas, the sum of roots is r1+r2=−24r_1 + r_2 = -24 and the product of roots is r1r2=−mr_1 r_2 = -m. Using the algebraic identity (r1−r2)2=(r1+r2)2−4r1r2(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1 r_2 with the given root difference ∣r1−r2∣=10|r_1 - r_2| = 10 yields 102=(−24)2−4(−m)10^2 = (-24)^2 - 4(-m), which simplifies to 100=576+4m100 = 576 + 4m. Subtracting 576 from both sides gives 4m=−4764m = -476, so m=−119m = -119.

Step-by-Step Solution

1
Find kk using the discriminant condition for equal roots of 3x2−kx+48=03x^2 - kx + 48 = 0.
k=24k = 24
A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has equal real roots when its discriminant b2−4ac=0b^2 - 4ac = 0. Here, (−k)2−4(3)(48)=0(-k)^2 - 4(3)(48) = 0, so k2=576k^2 = 576. Since k>0k > 0, k=24k = 24.
2
Set up the algebraic relationship for the root difference of x2+24x−m=0x^2 + 24x - m = 0.
(r1−r2)2=100(r_1 - r_2)^2 = 100
Given that the difference between the roots r1r_1 and r2r_2 is 10, ∣r1−r2∣=10|r_1 - r_2| = 10, which implies (r1−r2)2=100(r_1 - r_2)^2 = 100.
3
Apply Vieta's formulas to express (r1−r2)2(r_1 - r_2)^2 in terms of mm and solve.
m=−119m = -119
By Vieta's formulas for x2+24x−m=0x^2 + 24x - m = 0, the sum of roots is r1+r2=−24r_1 + r_2 = -24 and the product of roots is r1r2=−mr_1 r_2 = -m. Using the identity (r1−r2)2=(r1+r2)2−4r1r2(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1 r_2, substitute the known values: 100=(−24)2−4(−m)=576+4m100 = (-24)^2 - 4(-m) = 576 + 4m. Solving 576+4m=100576 + 4m = 100 gives 4m=−4764m = -476, so m=−119m = -119.

Key Concept

Discriminant analysis and root difference identity via Vieta's formulas
Question 15Question →

If the quadratic equation x2−kx+36=0x^2 - kx + 36 = 0 has two distinct positive integer roots, r1r_1 and r2r_2, and r1+2r_1 + 2 and r2+2r_2 + 2 are the roots of the quadratic equation x2−19x+m=0x^2 - 19x + m = 0, what is the value of mm?

Show answer & explanation

Answer: 70

Answer

The value of mm is 70.
By applying Vieta's formulas to both quadratic equations, we find that the sum of the original roots is r1+r2=15r_1 + r_2 = 15 and their product is r1r2=36r_1 r_2 = 36. Factoring x2−15x+36=(x−3)(x−12)=0x^2 - 15x + 36 = (x-3)(x-12) = 0 confirms the original roots are 33 and 1212. The new roots are 3+2=53+2=5 and 12+2=1412+2=14. Their product m=5×14=70m = 5 \times 14 = 70.

Step-by-Step Solution

1
Apply Vieta's relations to the first quadratic equation
r1+r2=kr_1 + r_2 = k and r1⋅r2=36r_1 \cdot r_2 = 36
For any quadratic equation x2+bx+c=0x^2 + bx + c = 0, the sum of roots is −b-b and the product of roots is cc.
2
Apply Vieta's relations to the second quadratic equation
(r1+2)+(r2+2)=19  ⟹  r1+r2=15(r_1 + 2) + (r_2 + 2) = 19 \implies r_1 + r_2 = 15
The coefficient of xx in x2−19x+m=0x^2 - 19x + m = 0 dictates that the sum of its roots equals 1919.
3
Determine the roots r1r_1 and r2r_2 and check integer constraints
r1=3r_1 = 3 and r2=12r_2 = 12
The integer factors of 3636 that sum to 1515 are 33 and 1212, satisfying all conditions.
4
Compute the constant term mm for the transformed equation
m=(3+2)(12+2)=5×14=70m = (3 + 2)(12 + 2) = 5 \times 14 = 70
The constant term mm equals the product of the transformed roots (r1+2)(r_1 + 2) and (r2+2)(r_2 + 2).

Key Concept

Vieta's Formulas and Quadratic Root Transformations
Estimated Time:2m 0s
Question 16Question →

Let SS be the set of all real solutions to the polynomial equation (x2−5x+6)(x2−7x+12)=(x−3)2(x−2)(x^2 - 5x + 6)(x^2 - 7x + 12) = (x - 3)^2(x - 2). What is the sum of all elements in set SS?

Show answer & explanation

Answer: 10

Answer

The sum of all distinct real elements in set SS is 10.
The correct answer is obtained by completely factoring both quadratic expressions on the left side into (x−2)(x−3)2(x−4)(x - 2)(x - 3)^2(x - 4). Moving all terms to one side gives (x−2)(x−3)2(x−4)−(x−3)2(x−2)=0(x - 2)(x - 3)^2(x - 4) - (x - 3)^2(x - 2) = 0. Factoring out (x−2)(x−3)2(x - 2)(x - 3)^2 yields (x−2)(x−3)2(x−5)=0(x - 2)(x - 3)^2(x - 5) = 0. The distinct solutions are 2, 3, and 5. Their sum is 10.

Step-by-Step Solution

1
Factor the quadratic expressions on the left side of the equation.
x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3) and x2−7x+12=(x−3)(x−4)x^2 - 7x + 12 = (x - 3)(x - 4). Substituting these into the left side yields (x−2)(x−3)2(x−4)(x - 2)(x - 3)^2(x - 4).
Factoring simplifies higher-degree polynomial expressions into linear factors.
2
Set up the equation with all terms on one side to avoid dividing by variable expressions.
(x−2)(x−3)2(x−4)−(x−3)2(x−2)=0(x - 2)(x - 3)^2(x - 4) - (x - 3)^2(x - 2) = 0.
Dividing by variable expressions can cause the loss of valid roots.
3
Factor out the common polynomial factor (x−2)(x−3)2(x - 2)(x - 3)^2.
(x−2)(x−3)2[(x−4)−1]=0(x - 2)(x - 3)^2 [(x - 4) - 1] = 0, which simplifies to (x−2)(x−3)2(x−5)=0(x - 2)(x - 3)^2(x - 5) = 0.
Factoring allows finding all roots by setting each factor equal to zero.
4
Identify the distinct real solutions forming set SS and compute their sum.
The roots are x=2x = 2, x=3x = 3, and x=5x = 5. Thus, S={2,3,5}S = \{2, 3, 5\}. The sum of elements in SS is 2+3+5=102 + 3 + 5 = 10.
A set contains only unique elements, so repeated roots are counted only once.

Key Concept

Factoring polynomial equations completely without dividing by variable expressions
Estimated Time:2m 0s
Question 17Question →

What is the sum of all real solutions to the equation x(x−3)=10x(x - 3) = 10?

Show answer & explanation

Answer: 3

Answer

The sum of all real solutions to the equation is 3.
Expanding the equation x(x−3)=10x(x - 3) = 10 yields x2−3x−10=0x^2 - 3x - 10 = 0. Factoring this expression gives (x−5)(x+2)=0(x - 5)(x + 2) = 0, yielding solutions x=5x = 5 and x=−2x = -2. The sum of these two real solutions is 5+(−2)=35 + (-2) = 3. Alternatively, applying Vieta's formula directly to x2−3x−10=0x^2 - 3x - 10 = 0 gives the sum of roots as −ba=−−31=3-\frac{b}{a} = -\frac{-3}{1} = 3.

Step-by-Step Solution

1
Expand the left side of the equation.
x2−3x=10x^2 - 3x = 10
Multiply xx across the binomial (x−3)(x - 3) to convert to standard polynomial form.
2
Rearrange the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−3x−10=0x^2 - 3x - 10 = 0
Subtract 10 from both sides so the equation equals zero.
3
Factor the quadratic expression.
(x−5)(x+2)=0(x - 5)(x + 2) = 0
Find two numbers that multiply to −10-10 and add up to −3-3, which are −5-5 and 22.
4
Solve for the roots of the equation.
x=5x = 5 or x=−2x = -2
Apply the zero-product property: set each factor equal to zero.
5
Calculate the sum of the solutions.
Sum =5+(−2)=3= 5 + (-2) = 3
Add the two real solutions together. Alternatively, by Vieta's formulas, the sum of roots is −ba=−−31=3-\frac{b}{a} = -\frac{-3}{1} = 3.

Key Concept

Quadratic Equations and Polynomial Factoring
Estimated Time:1m 0s
Question 18Question →

If 2 is a root of the quadratic equation x2−kx+24=0x^2 - kx + 24 = 0, where kk is a constant, and the quadratic equation x2−(k+2)x+m=0x^2 - (k + 2)x + m = 0 has exactly one real solution, what is the value of mm?

Show answer & explanation

Answer: 64

Answer

The value of mm is 64.
Substituting x=2x = 2 into x2−kx+24=0x^2 - kx + 24 = 0 gives 4−2k+24=04 - 2k + 24 = 0, which simplifies to 2k=282k = 28 or k=14k = 14. Substituting k=14k = 14 into x2−(k+2)x+m=0x^2 - (k + 2)x + m = 0 produces x2−16x+m=0x^2 - 16x + m = 0. For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 to have exactly one real solution, its discriminant b2−4acb^2 - 4ac must equal 0. Therefore, (−16)2−4(1)(m)=0(-16)^2 - 4(1)(m) = 0, which means 256−4m=0256 - 4m = 0, giving m=64m = 64.

Step-by-Step Solution

1
Substitute the known root x=2x = 2 into the equation x2−kx+24=0x^2 - kx + 24 = 0.
k=14k = 14
Since x=2x = 2 is a root of the quadratic equation, evaluating the expression at x=2x = 2 must equal zero.
2
Substitute k=14k = 14 into the second quadratic equation x2−(k+2)x+m=0x^2 - (k + 2)x + m = 0.
x2−16x+m=0x^2 - 16x + m = 0
This determines the linear coefficient of the second quadratic equation.
3
Set the discriminant of x2−16x+m=0x^2 - 16x + m = 0 to zero.
m=64m = 64
A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has exactly one real solution if and only if its discriminant b2−4acb^2 - 4ac equals zero.

Key Concept

Solving quadratic equations via root substitution and applying the discriminant condition for repeated roots.
Estimated Time:1m 30s
Question 19Question →

If 2x2+5x−3=02x^2 + 5x - 3 = 0 and x>0x > 0, what is the value of xx?

Show answer & explanation

Answer: 0.5

Answer

The value of xx is 0.5.
Factoring the quadratic equation 2x2+5x−3=02x^2 + 5x - 3 = 0 yields (2x−1)(x+3)=0(2x - 1)(x + 3) = 0. Setting each linear factor equal to zero gives two possible solutions for xx: x=0.5x = 0.5 and x=−3x = -3. Because the problem stipulates that x>0x > 0, the negative solution is discarded, leaving x=0.5x = 0.5.

Step-by-Step Solution

1
Factor the quadratic expression
(2x−1)(x+3)=0(2x - 1)(x + 3) = 0
Splitting the middle term 5x5x into 6x−x6x - x allows grouping to factor by grouping.
2
Find the roots of the equation
x=0.5x = 0.5 or x=−3x = -3
By the zero-product property, if the product of two factors is zero, at least one factor must be zero.
3
Apply the positivity constraint x>0x > 0
x=0.5x = 0.5
The root x=−3x = -3 violates the given condition that xx must be strictly greater than zero.

Key Concept

Solving quadratic equations by factoring and applying domain constraints
Question 20Question →

The quadratic equation x2−6x+a=0x^2 - 6x + a = 0 has two distinct real roots α\alpha and γ\gamma, and the quadratic equation y2−by+21=0y^2 - by + 21 = 0 has two distinct real roots β\beta and δ\delta. If α<β<γ<δ\alpha < \beta < \gamma < \delta and the four roots form an arithmetic progression in that order, what is the value of a+ba + b?

Show answer & explanation

Answer: 15

Answer

The value of a+ba + b is 15.
By representing the four ordered roots as r,r+d,r+2d,r+3dr, r+d, r+2d, r+3d, Vieta's formula for the sum of roots of the first equation yields r+(r+2d)=2(r+d)=6r + (r+2d) = 2(r+d) = 6, which implies r+d=3r+d = 3. Thus, the second root is β=3\beta = 3. Using the product of roots for the second equation, 3δ=213\delta = 21 gives δ=7\delta = 7. The common difference is d=(7−3)/2=2d = (7-3)/2 = 2, which gives the roots 1,3,5,71, 3, 5, 7. Finally, a=1×5=5a = 1 \times 5 = 5 and b=3+7=10b = 3 + 7 = 10, so a+b=15a + b = 15.

Step-by-Step Solution

1
Apply Vieta's formulas to both quadratic equations.
α+γ=6\alpha + \gamma = 6, αγ=a\alpha\gamma = a, β+δ=b\beta + \delta = b, and βδ=21\beta\delta = 21.
Vieta's relations connect the coefficients of a quadratic polynomial to the sum and product of its roots.
2
Set up the arithmetic progression representation for the roots.
α=r\alpha = r, β=r+d\beta = r + d, γ=r+2d\gamma = r + 2d, and δ=r+3d\delta = r + 3d, where d>0d > 0.
The roots form an increasing arithmetic progression in the order α,β,γ,δ\alpha, \beta, \gamma, \delta.
3
Substitute the expressions for α\alpha and γ\gamma into α+γ=6\alpha + \gamma = 6.
r+(r+2d)=2r+2d=2(r+d)=6  ⟹  r+d=3r + (r + 2d) = 2r + 2d = 2(r + d) = 6 \implies r + d = 3.
Combining terms simplifies the sum of the first and third terms of the arithmetic progression.
4
Identify the value of β\beta and solve for δ\delta.
β=r+d=3\beta = r + d = 3, so βδ=21  ⟹  3δ=21  ⟹  δ=7\beta\delta = 21 \implies 3\delta = 21 \implies \delta = 7.
Since β=r+d\beta = r + d, its value is directly determined as 3, allowing δ\delta to be solved from the product relation.
5
Calculate the common difference dd and the first term rr.
δ−β=2d=7−3=4  ⟹  d=2\delta - \beta = 2d = 7 - 3 = 4 \implies d = 2, and r=3−2=1r = 3 - 2 = 1.
The difference between the fourth and second terms of an AP is equal to 2d2d.
6
Find aa, bb, and their sum a+ba + b.
a=αγ=1×5=5a = \alpha\gamma = 1 \times 5 = 5, b=β+δ=3+7=10b = \beta + \delta = 3 + 7 = 10, so a+b=5+10=15a + b = 5 + 10 = 15.
With all four roots determined (1,3,5,71, 3, 5, 7), the missing coefficients are calculated using Vieta's formulas.

Key Concept

Combining Vieta's Formulas with Arithmetic Progressions to Solve Quadratic Systems
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