Question

Difficulty: MediumQuadratic Equations and Factoring

The quadratic equation x2kx+36=0x^2 - kx + 36 = 0, where kk is a positive constant, has two distinct real roots r1r_1 and r2r_2 such that r2r1=5r_2 - r_1 = 5. What is the value of kk?

Answer: 13

Answer

The value of kk is 13.
According to Vieta's formulas, for the quadratic equation x2kx+36=0x^2 - kx + 36 = 0, the sum of the roots is r1+r2=kr_1 + r_2 = k and the product of the roots is r1r2=36r_1 r_2 = 36. Using the identity (r2r1)2=(r1+r2)24r1r2(r_2 - r_1)^2 = (r_1 + r_2)^2 - 4r_1 r_2, we substitute the known values r2r1=5r_2 - r_1 = 5, r1+r2=kr_1 + r_2 = k, and r1r2=36r_1 r_2 = 36. This gives 52=k24(36)5^2 = k^2 - 4(36), which simplifies to 25=k214425 = k^2 - 144. Solving for k2k^2 gives k2=169k^2 = 169. Since kk is specified as a positive constant, k=13k = 13.

Step-by-Step Solution

1
Apply Vieta's formulas to the given quadratic equation
The sum of the roots is r1+r2=kr_1 + r_2 = k and the product of the roots is r1r2=36r_1 r_2 = 36.
For any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of roots is b/a-b/a and the product of roots is c/ac/a.
2
Relate the difference of the roots to their sum and product
(r2r1)2=(r1+r2)24r1r2(r_2 - r_1)^2 = (r_1 + r_2)^2 - 4r_1 r_2
Expanding both sides shows that r222r1r2+r12=r12+2r1r2+r224r1r2r_2^2 - 2r_1 r_2 + r_1^2 = r_1^2 + 2r_1 r_2 + r_2^2 - 4r_1 r_2, which is an algebraic identity.
3
Substitute the known values into the identity
52=k24(36)    25=k21445^2 = k^2 - 4(36) \implies 25 = k^2 - 144
We are given that r2r1=5r_2 - r_1 = 5, r1r2=36r_1 r_2 = 36, and r1+r2=kr_1 + r_2 = k.
4
Solve for the positive constant kk
k2=169    k=13k^2 = 169 \implies k = 13
Adding 144 to both sides gives k2=169k^2 = 169. Taking the positive square root because k>0k > 0 yields k=13k = 13.

Key Concept

Vieta's Formulas and Root Difference Identity
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