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Difficulty: Very hardQuadratic Equations

For a certain constant aa, the quadratic equation 3x2−2ax+a−4=03x^2 - 2ax + a - 4 = 0 has real roots x1x_1 and x2x_2. If x12+x22=289x_1^2 + x_2^2 = \frac{28}{9}, what is the sum of all possible values of aa?

  1. A
    34\frac{3}{4}
  2. 32\frac{3}{2}Answer
  3. C
    −32-\frac{3}{2}
  4. D
    3

Answer

The sum of all possible values of aa is \frac{3}{2}.
The correct answer is derived by expressing the sum of the squares of the roots, x12+x22x_1^2 + x_2^2, as (x1+x2)2−2x1x2(x_1 + x_2)^2 - 2x_1x_2. Applying Vieta's formulas to the equation 3x2−2ax+a−4=03x^2 - 2ax + a - 4 = 0 yields x1+x2=2a3x_1 + x_2 = \frac{2a}{3} and x1x2=a−43x_1 x_2 = \frac{a - 4}{3}. Substituting these values into the identity gives 4a29−2(a−4)3=289\frac{4a^2}{9} - \frac{2(a - 4)}{3} = \frac{28}{9}. Multiplying by 9 and simplifying leads to the quadratic equation 2a2−3a−2=02a^2 - 3a - 2 = 0. Since this equation has real solutions and the original equation always has real roots for any real aa, the sum of all possible values of aa is given by the negative linear coefficient divided by the leading coefficient, which is \frac{3}{2}.

Step-by-Step Solution

1
Express the sum of squares x12+x22x_1^2 + x_2^2 in terms of the sum and product of the roots.
x12+x22=(x1+x2)2−2x1x2x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2
This allows the use of Vieta's formulas to relate the roots of the quadratic equation to its coefficients.
2
Apply Vieta's formulas to the quadratic equation 3x2−2ax+a−4=03x^2 - 2ax + a - 4 = 0 to find the sum and product of the roots.
x1+x2=2a3x_1 + x_2 = \frac{2a}{3} and x1x2=a−43x_1 x_2 = \frac{a - 4}{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 is CA\frac{C}{A}.
3
Substitute the expressions for the sum and product of the roots into the identity from Step 1 and set it equal to the given value.
(2a3)2−2(a−43)=289\left(\frac{2a}{3}\right)^2 - 2\left(\frac{a - 4}{3}\right) = \frac{28}{9}
This sets up a solvable equation for the parameter aa.
4
Simplify the algebraic equation by multiplying both sides by 9 and expanding terms.
4a2−6(a−4)=28  ⟹  4a2−6a−4=0  ⟹  2a2−3a−2=04a^2 - 6(a - 4) = 28 \implies 4a^2 - 6a - 4 = 0 \implies 2a^2 - 3a - 2 = 0
Eliminating denominators and grouping terms yields a standard form quadratic equation in aa.
5
Verify that the solutions for aa are real and that they yield real roots for xx in the original equation.
For 2a2−3a−2=02a^2 - 3a - 2 = 0, the discriminant Δa=25>0\Delta_a = 25 > 0, so aa has real solutions. The discriminant of the original equation is Δx=4a2−12a+48\Delta_x = 4a^2 - 12a + 48, which is positive for all real values of aa since its own discriminant is negative.
This confirms that the solutions for aa are valid and satisfy the condition that x1x_1 and x2x_2 are real numbers.
6
Use Vieta's formula on the quadratic equation 2a2−3a−2=02a^2 - 3a - 2 = 0 to find the sum of all possible values of aa.
Sum of values =−−32=32= -\frac{-3}{2} = \frac{3}{2}
The sum of the roots of a quadratic equation in the form Aa2+Ba+C=0Aa^2 + Ba + C = 0 is equal to −BA-\frac{B}{A}.

Key Concept

Vieta's Formulas and Algebraic Identities for Quadratic Equations
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Topics

Quadratic Equations

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