If x = 5 is a solution to the quadratic equation x^2 - (k + 3)x + 3k + 1 = 0, where k is a… — GRE General Test

If $x = 5$ is a solution to the quadratic equation $x^2 - (k + 3)x + 3k + 1 = 0$ , where $k$ is a constant, what is the value of the other solution?

\frac{7}{2}

Substituting

x = 5

into

x^2 - (k + 3)x + 3k + 1 = 0

yields

25 - 5(k + 3) + 3k + 1 = 0

, which simplifies to

11 - 2k = 0

, so

k = \frac{11}{2}

. Substituting

k = \frac{11}{2}

back into the product of roots formula

x_1 \cdot x_2 = 3k + 1

gives

5 \cdot x_2 = 3\left(\frac{11}{2}\right) + 1 = \frac{35}{2}

. Dividing by

5

yields the second root

x_2 = \frac{7}{2}

Substitute the known solution

x = 5

into the quadratic equation to solve for

k

5^2 - (k + 3)(5) + 3k + 1 = 0 \implies 25 - 5k - 15 + 3k + 1 = 0 \implies 11 - 2k = 0 \implies k = \frac{11}{2}

Since

x = 5

is a solution, it must satisfy the equation.

Use Vieta's formulas to find the other solution

x_2

Product of roots

x_1 \cdot x_2 = 3k + 1 = 3\left(\frac{11}{2}\right) + 1 = \frac{35}{2}

. Since

x_1 = 5

5 \cdot x_2 = \frac{35}{2} \implies x_2 = \frac{7}{2}

By Vieta's formulas for a standard quadratic

x^2 + bx + c = 0

, the product of the roots equals

c

Quadratic equations, Vieta's formulas, and parameter evaluation

Question