Question

If x = (3 + √7) /2, find the value of 4x² + (1/x²) .

Answer 1

Explanation:

here x is root so other root will be conj. pair of given value

get a quad. eqn in x

divide the given eqn

Answer 2

The value of 4x² + (1/x²) is 24 (appox).

Explanation:

The value of the x given in the question is:

$\bold{x=\left(\frac{3+\sqrt{7}}{2}\right)}$

Now, we need to find the square value of x.

$\bold{x^{2}=\left(\frac{3+\sqrt{7}}{2}\right)\left(\frac{3+\sqrt{7}}{2}\right)}$

$\bold{x^{2}=\frac{9+3 \sqrt{7}+3 \sqrt{7}+7}{4}}$

$\bold{x^{2}=\frac{9+6 \sqrt{7}+7}{4}}$

$\bold{x^{2}=\frac{16+6 \sqrt{7}}{4}}$

$\bold{x^{2}=\frac{2(8+3 \sqrt{7})}{4}}$

$\bold{x^{2}=\frac{8+3 \sqrt{7}}{2}}$

On substituting the value of x in given equation, we get,

$\bold{4 x^{2}+\frac{1}{x^{2}}=4\left(\frac{8+3 \sqrt{7}}{2}\right)+\frac{1}{\frac{8+3 \sqrt{7}}{2}}}$

$\bold{4 x^{2}+\frac{1}{x^{2}}=2(8+3 \sqrt{7})+\frac{2}{8+3 \sqrt{7}}}$

$\bold{4 x^{2}+\frac{1}{x^{2}}=\frac{2(8+3 \sqrt{7})^{2}+2}{8+3 \sqrt{7}}}$

$\bold{4 x^{2}+\frac{1}{x^{2}}=\frac{2(64+63+24 \sqrt{7})+2}{8+3 \sqrt{7}}}$

$\bold{4 x^{2}+\frac{1}{x^{2}}=\frac{128+126+48 \sqrt{7}+2}{8+3 \sqrt{7}}}$

$\bold{4 x^{2}+\frac{1}{x^{2}}=\frac{256+48 \sqrt{7}}{8+3 \sqrt{7}}}$

The value of √7 is 2.65

$\bold{4 x^{2}+\frac{1}{x^{2}}=\frac{256+48(2.65)}{8+3(2.65)}}$

$\bold{4 x^{2}+\frac{1}{x^{2}}=\frac{256+127.2}{8+7.95}}$

$\bold{4 x^{2}+\frac{1}{x^{2}}=\frac{383.2}{15.95}}$

$\bold{\therefore 4 x^{2}+\frac{1}{x^{2}}=24.02 \approx 24}$