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Difficulty: HardQuadratic Equations and Factoring

For a nonzero real constant mm, the quadratic equation x2−(m+6m)x+6=0x^2 - \left(m + \frac{6}{m}\right)x + 6 = 0 has real roots aa and bb. If a2+b2=25a^2 + b^2 = 25, what is the sum of all possible values of m2m^2?

  1. A
    1313
  2. 2525Answer
  3. C
    3737
  4. D
    4949
  5. E
    55

Answer

The sum of all possible values of m2m^2 is 2525.
Using Vieta's formulas on x2−(m+6m)x+6=0x^2 - \left(m + \frac{6}{m}\right)x + 6 = 0, the sum of the roots is a+b=m+6ma + b = m + \frac{6}{m} and the product is ab=6ab = 6. Substituting these into the identity a2+b2=(a+b)2−2ab=25a^2 + b^2 = (a+b)^2 - 2ab = 25 yields (m+6m)2−12=25\left(m + \frac{6}{m}\right)^2 - 12 = 25, which simplifies to m2+36m2=25m^2 + \frac{36}{m^2} = 25. Setting u=m2u = m^2 gives u2−25u+36=0u^2 - 25u + 36 = 0. Since this quadratic has two distinct positive real roots for uu, the sum of all possible values of m2m^2 is given by Vieta's formula as 2525.

Step-by-Step Solution

1
Apply Vieta's formulas to relate the roots aa and bb to the coefficients of the given quadratic equation.
a+b=m+6ma + b = m + \frac{6}{m} and ab=6ab = 6.
For any quadratic equation x2−Sx+P=0x^2 - Sx + P = 0, the sum of the roots is SS and the product of the roots is PP.
2
Express a2+b2a^2 + b^2 using the algebraic identity for the sum of squares.
a2+b2=(a+b)2−2ab=(m+6m)2−2(6)=(m+6m)2−12a^2 + b^2 = (a + b)^2 - 2ab = \left(m + \frac{6}{m}\right)^2 - 2(6) = \left(m + \frac{6}{m}\right)^2 - 12.
Rewriting the sum of squares allows substitution of the known sum and product of the roots.
3
Set the expression equal to the given value 2525 and expand the square.
m2+2(m)(6m)+36m2−12=25  ⟹  m2+12+36m2−12=25  ⟹  m2+36m2=25m^2 + 2(m)\left(\frac{6}{m}\right) + \frac{36}{m^2} - 12 = 25 \implies m^2 + 12 + \frac{36}{m^2} - 12 = 25 \implies m^2 + \frac{36}{m^2} = 25.
Expanding (m+6m)2\left(m + \frac{6}{m}\right)^2 yields m2+12+36m2m^2 + 12 + \frac{36}{m^2}, which simplifies directly after subtracting 1212.
4
Formulate a quadratic equation in terms of u=m2u = m^2.
u+36u=25  ⟹  u2−25u+36=0u + \frac{36}{u} = 25 \implies u^2 - 25u + 36 = 0.
Multiplying through by u=m2u = m^2 transforms the rational equation into a standard polynomial quadratic equation.
5
Find the sum of all valid solutions for u=m2u = m^2 using Vieta's formulas.
The discriminant of u2−25u+36=0u^2 - 25u + 36 = 0 is Δ=(−25)2−4(1)(36)=625−144=481>0\Delta = (-25)^2 - 4(1)(36) = 625 - 144 = 481 > 0, and the product of the roots is 36>036 > 0, so both roots u1,u2u_1, u_2 are positive real numbers. By Vieta's formulas, u1+u2=25u_1 + u_2 = 25.
Since both roots u1u_1 and u2u_2 are positive real numbers, they correspond to valid real values of m2m^2, making their sum equal to 2525.

Key Concept

Quadratic Equations, Vieta's Formulas, and Factoring Substitution
Estimated Time:2m 0s
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Topics

Quadratic Equations and Factoring

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