For a constant a > 0, the quadratic equation x^2 - ax + (2a + 1) = 0 has two real roots, r… — GMAT

For a constant $a > 0$ , the quadratic equation $x^2 - ax + (2a + 1) = 0$ has two real roots, $r$ and $s$ . If $r^2 + s^2 = 43$ , what is the value of $a$ ?

Answer: 9

Answer

The value of

a

is 9.

According to Vieta's formulas, the sum of the roots of

x^2 - ax + (2a + 1) = 0

r + s = a

and the product of the roots is

rs = 2a + 1

. Using the identity

r^2 + s^2 = (r + s)^2 - 2rs

, substitute the Vieta expressions to obtain

r^2 + s^2 = a^2 - 2(2a + 1) = a^2 - 4a - 2

. Setting this equal to 43 yields

a^2 - 4a - 2 = 43

, which simplifies to

a^2 - 4a - 45 = 0

. Factoring the quadratic gives

(a - 9)(a + 5) = 0

, giving solutions

a = 9

a = -5

. Because the problem specifies that

a > 0

a

must be 9. Checking the discriminant

\Delta = (-9)^2 - 4(1)(19) = 5 > 0

confirms that two real roots exist.

Step-by-Step Solution

Express the sum and product of the roots in terms of

a

using Vieta's formulas.

r + s = a

and

rs = 2a + 1

For a standard quadratic equation

x^2 + bx + c = 0

, the sum of the roots is

-b

and the product of the roots is

c

Relate

r^2 + s^2

(r + s)

and

rs

r^2 + s^2 = (r + s)^2 - 2rs = a^2 - 2(2a + 1) = a^2 - 4a - 2

Expanding

(r + s)^2 = r^2 + 2rs + s^2

allows expressing

r^2 + s^2

in terms of known quantities.

Substitute

r^2 + s^2 = 43

into the equation and solve for

a

a^2 - 4a - 2 = 43 \implies a^2 - 4a - 45 = 0 \implies (a - 9)(a + 5) = 0

Rearranging terms forms a new quadratic equation in terms of

a

Apply the positivity constraint

a > 0

and verify that the roots are real.

Since

a > 0

a = 9

. The discriminant of the original equation is

\Delta = (-9)^2 - 4(1)(19) = 5 > 0

, confirming real roots exist.

The question specifies that

a

is positive and that

r

and

s

are real numbers.

Key Concept

Vieta's Formulas and Quadratic Modeling

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