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Difficulty: MediumQuadratic Equations

In the quadratic equation 3x2−18x+c=03x^2 - 18x + c = 0, cc is a constant. If the sum of the squares of the two real solutions to this equation is 2626, what is the value of cc?

Answer: 15

Answer

The value of cc is 1515.
To find the value of cc, we apply Vieta's formulas to the equation 3x2−18x+c=03x^2 - 18x + c = 0. The sum of the solutions is x1+x2=−−183=6x_1 + x_2 = -\frac{-18}{3} = 6, and the product of the solutions is x1x2=c3x_1 x_2 = \frac{c}{3}. Using the identity x12+x22=(x1+x2)2−2x1x2x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2, we substitute the given values: 26=62−2(c3)26 = 6^2 - 2\left(\frac{c}{3}\right). This simplifies to 26=36−2c326 = 36 - \frac{2c}{3}. Subtracting 3636 from both sides yields −10=−2c3-10 = -\frac{2c}{3}, which gives 2c=302c = 30, and thus c=15c = 15. Alternatively, since the vertex of the corresponding parabola is at x=−−182(3)=3x = -\frac{-18}{2(3)} = 3, the two solutions can be represented symmetrically as 3+d3 + d and 3−d3 - d. The sum of their squares is (3+d)2+(3−d)2=(9+6d+d2)+(9−6d+d2)=18+2d2(3 + d)^2 + (3 - d)^2 = (9 + 6d + d^2) + (9 - 6d + d^2) = 18 + 2d^2. Setting this equal to the given value of 2626 yields 18+2d2=26  ⟹  2d2=8  ⟹  d2=4  ⟹  d=218 + 2d^2 = 26 \implies 2d^2 = 8 \implies d^2 = 4 \implies d = 2. Therefore, the solutions are 3+2=53 + 2 = 5 and 3−2=13 - 2 = 1. Substituting either solution back into the original equation, such as x=1x = 1, yields 3(1)2−18(1)+c=0  ⟹  −15+c=0  ⟹  c=153(1)^2 - 18(1) + c = 0 \implies -15 + c = 0 \implies c = 15.

Step-by-Step Solution

1
Find the sum and product of the solutions to the quadratic equation.
The sum of the solutions is x1+x2=6x_1 + x_2 = 6, and their product is x1x2=c3x_1 x_2 = \frac{c}{3}.
According to Vieta's formulas, for a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the solutions is −ba-\frac{b}{a} and the product of the solutions is ca\frac{c}{a}.
2
Relate the sum of the squares of the solutions to their sum and product using the algebraic identity.
x12+x22=(x1+x2)2−2x1x2  ⟹  26=(6)2−2(c3)x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 \implies 26 = (6)^2 - 2\left(\frac{c}{3}\right).
The algebraic identity (x1+x2)2=x12+2x1x2+x22(x_1 + x_2)^2 = x_1^2 + 2x_1 x_2 + x_2^2 can be rearranged to express the sum of squares x12+x22x_1^2 + x_2^2 in terms of the sum and product.
3
Solve the resulting linear equation for the constant cc.
26=36−2c3  ⟹  2c3=10  ⟹  c=1526 = 36 - \frac{2c}{3} \implies \frac{2c}{3} = 10 \implies c = 15.
Subtracting 3636 from both sides gives −10=−2c3-10 = -\frac{2c}{3}. Multiplying by −3-3 gives 30=2c30 = 2c, which yields c=15c = 15.

Key Concept

Vieta's Formulas and Algebraic Identities
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Topics

Quadratic Equations

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