If x is a real number such that x^2 - 5x + 3 = 0, what is the value of x^2 + \frac{9}{x^2}… — GMAT

If $x$ is a real number such that $x^2 - 5x + 3 = 0$ , what is the value of $x^2 + \frac{9}{x^2}$ ?

Answer: 19

The value of

x^2 + \frac{9}{x^2}

is 19.

Dividing

x^2 - 5x + 3 = 0

x

gives

x + \frac{3}{x} = 5

. Squaring both sides yields

\left(x + \frac{3}{x}\right)^2 = x^2 + 2(x)\left(\frac{3}{x}\right) + \frac{9}{x^2} = 25

, which simplifies to

x^2 + 6 + \frac{9}{x^2} = 25

. Subtracting 6 from both sides yields the final answer of 19.

Divide the quadratic equation by

x

x + \frac{3}{x} = 5

Since

x = 0

is not a root of

x^2 - 5x + 3 = 0

, dividing by

x

rearranges the terms into

x

and its reciprocal component.

Square both sides of

x + \frac{3}{x} = 5

x^2 + 6 + \frac{9}{x^2} = 25

Applying the algebraic identity

(a + b)^2 = a^2 + 2ab + b^2

generates the required expression

x^2 + \frac{9}{x^2}

alongside a constant middle term.

Isolate

x^2 + \frac{9}{x^2}

x^2 + \frac{9}{x^2} = 19

Subtracting the constant cross-term 6 from 25 yields the target value.

Algebraic manipulation of quadratic equations using reciprocal square identities

Estimated Time:2m 0s

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