In the quadratic equation x^2 - kx + 5 = 0, k is a constant. If the solutions to the equat… — SAT

In the quadratic equation $x^2 - kx + 5 = 0$ , $k$ is a constant. If the solutions to the equation are $x = 3 \pm \sqrt{a}$ , where $a$ is a positive constant, what is the value of $a$ ?

Answer: 4

Answer

The correct answer is 4. By relating the roots of a quadratic equation to its coefficients, the sum of the roots is

(3 + \sqrt{a}) + (3 - \sqrt{a}) = 6

, which equals the linear coefficient term

k

x^2 - kx + 5 = 0

. Using

k = 6

, we can rewrite the equation as

x^2 - 6x + 5 = 0

. Completing the square yields

(x - 3)^2 - 9 + 5 = 0

, which simplifies to

(x - 3)^2 = 4

, or

x = 3 \pm \sqrt{4}

. Comparing this to the given expression

x = 3 \pm \sqrt{a}

gives

a = 4

. Alternatively, using the product of roots,

(3 + \sqrt{a})(3 - \sqrt{a}) = 9 - a

. Since the product of roots is the constant term 5, we have

9 - a = 5

, which directly yields

a = 4

Step-by-Step Solution

Find the sum of the given solutions to determine the value of the constant

k

k = 6

The sum of the solutions

(3 + \sqrt{a}) + (3 - \sqrt{a})

6

. For a quadratic equation in the form

x^2 - kx + c = 0

, the sum of the solutions is equal to

k

Substitute

k = 6

back into the quadratic equation.

x^2 - 6x + 5 = 0

This gives the complete quadratic equation with numerical coefficients.

Solve the quadratic equation by completing the square to express the solutions in the form

3 \pm \sqrt{a}

x = 3 \pm \sqrt{4}

Completing the square gives

(x - 3)^2 - 9 + 5 = 0

, which simplifies to

(x - 3)^2 = 4

, so

x - 3 = \pm \sqrt{4}

Compare the obtained solutions to the given form

x = 3 \pm \sqrt{a}

to identify the value of

a

a = 4

Comparing

x = 3 \pm \sqrt{4}

with

x = 3 \pm \sqrt{a}

yields

a = 4

Key Concept

Relationship between roots and coefficients of a quadratic equation, or solving quadratic equations by completing the square.

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