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Difficulty: EasyQuadratic Equations

The quadratic equation 3x2+12x−15=03x^2 + 12x - 15 = 0 has solutions rr and ss, where r>sr > s. What is the value of r−sr - s?

  1. A
    4
  2. 6Answer
  3. C
    -4
  4. D
    -6

Answer

6
The correct answer is 6. Dividing the given quadratic equation 3x2+12x−15=03x^2 + 12x - 15 = 0 by 3 simplifies it to x2+4x−5=0x^2 + 4x - 5 = 0. Factoring this expression gives (x+5)(x−1)=0(x + 5)(x - 1) = 0, which yields the solutions x=1x = 1 and x=−5x = -5. Since we are given that r>sr > s, we define r=1r = 1 and s=−5s = -5. The difference between the roots is r−s=1−(−5)=1+5=6r - s = 1 - (-5) = 1 + 5 = 6.

Step-by-Step Solution

1
Divide the entire equation 3x2+12x−15=03x^2 + 12x - 15 = 0 by 3 to simplify it.
x2+4x−5=0x^2 + 4x - 5 = 0
Simplifying the equation makes it easier to factor by reducing the coefficients.
2
Factor the simplified quadratic equation x2+4x−5=0x^2 + 4x - 5 = 0.
(x+5)(x−1)=0(x + 5)(x - 1) = 0
We look for two numbers that multiply to -5 and add to 4, which are 5 and -1.
3
Solve for the roots by setting each factor to zero, and assign the variables rr and ss such that r>sr > s.
x=1x = 1 and x=−5x = -5, meaning r=1r = 1 and s=−5s = -5.
Setting the factors to zero gives the solutions. Since the problem specifies r > s, the larger solution (1) is assigned to r and the smaller solution (-5) is assigned to s.
4
Calculate the value of r−sr - s by substituting the values of rr and ss.
r−s=1−(−5)=6r - s = 1 - (-5) = 6
Subtracting s from r gives the final value of 6.

Key Concept

Solving quadratic equations by factoring and finding the difference between roots.

Alternative Method

Alternatively, Vieta's formulas can be used. For any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is r+s=−b/ar + s = -b/a and the product of the roots is rs=c/ars = c/a. For 3x2+12x−15=03x^2 + 12x - 15 = 0, we find r+s=−12/3=−4r + s = -12/3 = -4 and rs=−15/3=−5rs = -15/3 = -5. The relationship between the sum, product, and difference of two numbers is given by the algebraic identity (r−s)2=(r+s)2−4rs(r - s)^2 = (r + s)^2 - 4rs. Substituting our values gives (r−s)2=(−4)2−4(−5)=16+20=36(r - s)^2 = (-4)^2 - 4(-5) = 16 + 20 = 36. Since r>sr > s, the difference r−sr - s must be positive, so we take the positive square root: r−s=36=6r - s = \sqrt{36} = 6.
Estimated Time:1m 0s
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Topics

Quadratic Equations

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