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Difficulty: MediumQuadratic Equations and Factoring

If mm and nn are the two distinct real solutions to the quadratic equation 2x2−8x+c=02x^2 - 8x + c = 0, where cc is a constant, and m2+n2=12m^2 + n^2 = 12, what is the value of cc?

  1. A
    -4
  2. 4Answer
  3. C
    8
  4. D
    2
  5. E
    26

Answer

The value of cc is 44.
By Vieta's formulas for the quadratic equation 2x2−8x+c=02x^2 - 8x + c = 0, the sum of the roots is m+n=4m + n = 4 and the product of the roots is mn=c/2mn = c/2. Squaring the sum gives (m+n)2=m2+2mn+n2=16(m + n)^2 = m^2 + 2mn + n^2 = 16. Substituting m2+n2=12m^2 + n^2 = 12 gives 12+2(c/2)=1612 + 2(c/2) = 16, which simplifies to 12+c=1612 + c = 16, so c=4c = 4.

Step-by-Step Solution

1
Apply Vieta's formulas to determine the sum and product of the roots in terms of the given quadratic coefficients.
For 2x2−8x+c=02x^2 - 8x + c = 0, the sum of roots is m+n=−−82=4m + n = -\frac{-8}{2} = 4, and the product of roots is m⋅n=c2m \cdot n = \frac{c}{2}.
For a 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 m2+n2m^2 + n^2 to the sum and product of roots using the identity (m+n)2=m2+n2+2mn(m + n)^2 = m^2 + n^2 + 2mn.
m2+n2=(m+n)2−2mnm^2 + n^2 = (m + n)^2 - 2mn
Rewriting the sum of squares in terms of (m+n)(m+n) and mnmn allows substitution of the known values.
3
Substitute m+n=4m + n = 4, mn=c2mn = \frac{c}{2}, and m2+n2=12m^2 + n^2 = 12 into the equation and solve for cc.
12=(4)2−2(c2)  ⟹  12=16−c  ⟹  c=412 = (4)^2 - 2\left(\frac{c}{2}\right) \implies 12 = 16 - c \implies c = 4.
Direct algebraic simplification yields the constant cc.

Key Concept

Relationship between roots and coefficients of a quadratic equation (Vieta's Formulas) and algebraic identities.
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Topics

Quadratic Equations and Factoring

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