If x is a real number such that x^2 - 5x + 1 = 0, what is the value of x^2 + \frac{1}{x^2}… — GMAT

If $x$ is a real number such that $x^2 - 5x + 1 = 0$ , what is the value of $x^2 + \frac{1}{x^2}$ ?

Cevap: 23

The value of

x^2 + \frac{1}{x^2}

is 23.

Dividing the quadratic equation

x^2 - 5x + 1 = 0

x

yields

x + \frac{1}{x} = 5

. Squaring both sides of this identity gives

\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} = 25

. Subtracting 2 from both sides results in

x^2 + \frac{1}{x^2} = 23

Divide the given quadratic equation

x^2 - 5x + 1 = 0

x

x - 5 + \frac{1}{x} = 0 \implies x + \frac{1}{x} = 5

Since

x \neq 0

, dividing by

x

isolates the sum of

x

and its reciprocal.

Square both sides of the expression

x + \frac{1}{x} = 5

\left(x + \frac{1}{x}\right)^2 = 25 \implies x^2 + 2 + \frac{1}{x^2} = 25

Expanding the binomial square produces the required quadratic sum along with a constant cross-term.

Isolate

x^2 + \frac{1}{x^2}

by subtracting 2 from 25

x^2 + \frac{1}{x^2} = 23

Subtracting the constant cross-term yields the exact requested numeric value.

Algebraic transformation of quadratic equations into reciprocal power sums

Soru