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Difficulty: MediumQuadratic Equations and Factoring

If the quadratic equation 3x2−12x+c=03x^2 - 12x + c = 0 has two real roots, r1r_1 and r2r_2, such that r12+r22=10r_1^2 + r_2^2 = 10, what is the value of the constant cc?

  1. A
    3
  2. B
    6
  3. 9Answer
  4. D
    18
  5. E
    -9

Answer

The value of the constant cc is 9.
The correct answer is 9. By Vieta's formulas, the sum of the roots is r1+r2=−−123=4r_1 + r_2 = -\frac{-12}{3} = 4 and the product of the roots is r1r2=c3r_1 r_2 = \frac{c}{3}. Using the identity r12+r22=(r1+r2)2−2r1r2r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2, we substitute the given values to get 10=16−2c310 = 16 - \frac{2c}{3}, which simplifies to 2c3=6\frac{2c}{3} = 6, giving c=9c = 9.

Step-by-Step Solution

1
Apply Vieta's formulas to express the sum and product of the roots in terms of equation coefficients.
r1+r2=−−123=4r_1 + r_2 = -\frac{-12}{3} = 4 and r1r2=c3r_1 r_2 = \frac{c}{3}.
For any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is −ba-\frac{b}{a} and the product of the roots is ca\frac{c}{a}.
2
Relate the sum of squares r12+r22r_1^2 + r_2^2 to (r1+r2)(r_1 + r_2) and r1r2r_1 r_2.
r12+r22=(r1+r2)2−2r1r2r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2.
Expanding (r1+r2)2=r12+2r1r2+r22(r_1 + r_2)^2 = r_1^2 + 2r_1 r_2 + r_2^2 and rearranging gives the identity for the sum of squares.
3
Substitute the known values into the identity and solve for cc.
10=42−2(c3)  ⟹  10=16−2c3  ⟹  2c3=6  ⟹  c=910 = 4^2 - 2\left(\frac{c}{3}\right) \implies 10 = 16 - \frac{2c}{3} \implies \frac{2c}{3} = 6 \implies c = 9.
Substituting r12+r22=10r_1^2 + r_2^2 = 10 and r1+r2=4r_1 + r_2 = 4 isolates the single variable cc.

Key Concept

Vieta's Formulas and Algebraic Identities for Quadratic Equations

Alternative Method

Alternatively, factor 3(x2−4x+3)=03(x^2 - 4x + 3) = 0 directly once r1r_1 and r2r_2 are identified. Since r1+r2=4r_1 + r_2 = 4 and r12+r22=10r_1^2 + r_2^2 = 10, solving the system of equations for r1r_1 and r2r_2 yields roots of 1 and 3. The product of these roots is (1)(3)=3(1)(3) = 3, so c3=3\frac{c}{3} = 3, which gives c=9c = 9.
Estimated Time:1m 30s
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Topics

Quadratic Equations and Factoring

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