Question

5. If 2x = 3 + root 7, find the value of : 4x ^2 + 1/x^2 ​

Answer 1

$\bold \red{The \: value \: of \: 4x^{2} +\frac{1}{x^{2} }4x} \\ is \: \frac{256+96\sqrt{7} }{8+3\sqrt{7} }$

Step-by-step explanation:

$\bold \red{Given, 2x = 3 + \sqrt{7} }$

Squaring both sides,

$(2x)^{2} = (3+\sqrt{7} )^{2}$

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

$[(a+b)^{2} = a^{2} +b^{2} + 2ab]$

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

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

Now,

$\bold \red{\begin{gathered}4x^{2} + \frac{1}{x^{2} } \\= (16+6\sqrt{7} ) + \frac{1}{\frac{(16+6\sqrt{7}) }{4}}\end{gathered} }$

[Using equation(i) & (ii)]

$\bold \red{\begin{gathered}= (16 + 6\sqrt{7} ) +\frac{4}{(16+6\sqrt{7}) } \\= \frac{(16+6\sqrt{7})^{2} +4 }{16+6\sqrt{7}} \\= \frac{256 + 252 + 192\sqrt{7}+4 }{16+6\sqrt7} \\= \frac{512+192\sqrt{7} }{16+6\sqrt{7}} \\=\frac{256+96\sqrt{7}}{8+3\sqrt{7}} \\\end{gathered} }$

$\bold \red{Hence, the \: value \: of \: 4x^{2} +\frac{1}{x^{2} }}$

$\bold \red{Hence, the \: value \: of \: 4x^{2} +\frac{1}{x^{2} }} \\ \bold \red{is \: \frac{256+96\sqrt{7} }{8+3\sqrt{7} } }$

.

Answer 2

Answer:

Given:-

If 2x = 3 + √7, find the value of : $4x² + \dfrac{1}{x²}$ .

To Find:-

The value of : $4x² + \dfrac{1}{x²}$.

Note:-

●》Here from the Given question, we will find the value of "x" and then we will apply the "x" value to solve it.

●》For finding unknown values, known values needs be transposed from its side to another and signs are also changed or not ( signs are not changed in multiple and divisional value ). For example - Multiple becomes divisional.

●》In a fraction, if there is 1 numerator and 2 denominators then we should reciprocated it for calculations and in this divisional becomes multiple. For example - $\dfrac{1}{\frac{2}{3}} \implies \dfrac{1}{2} × 3$

Formula Used:-

■》( a + b )² = a² + b² + 2ab.

Solution:-

[ Note first point ]

$\huge\red{2x = 3 + √7}$

$\huge\red{\ \ \ \ x = ?}$

☆ So~

▪︎$2x = 3 + √7$

☆ Squaring both side~

▪︎$( 2x )² = ( 3 + √7 )²$

According to Formula Used and by square, root will be canceled~

▪︎$4x² = 3² + 7 + 2 × 3 × √7$

▪︎$4x² = 9 + 7 + 6 × √7$

▪︎$4x² = 16 + 6 × √7$

▪︎$4x² = 16 + 6√7$

According to note second point ( transposing )~

▪︎$x² = \dfrac{16 + 6√7}{4}$

▪︎$x = \sqrt\dfrac{16 + 6√7}{4}$

$\huge\pink{x = \sqrt\dfrac{16 + 6√7}{4}}$

______________________________________

$\huge\red{x = \sqrt\dfrac{16 + 6√7}{4}}$

$\huge\red{Value \ \ of \ \ 4x² + \dfrac{1}{x²} = ?}$

☆ According to note first point only~

▪︎$4x² + \dfrac{1}{x²}$

▪︎$4 × ( \sqrt\dfrac{16 + 6√7}{4} )² + \dfrac{1}{( \sqrt\frac{16 + 6√7}{4} )²}$

☆ Roots will be canceled by square and other numbers will be multiplied by square~

▪︎$4 × \dfrac{16 + 6√7}{4} + \dfrac{1}{\frac{16 + 6√7}{4}}$

According to note third point~

▪︎$4 × \dfrac{16 + 6√7}{4} + \dfrac{1}{16 + 6√7} × 4$

▪︎$1 × 16 + 6√7 + \dfrac{4}{16 + 6√7}$

▪︎$16 + 6√7 + \dfrac{4}{16 + 6√7}$

☆ L.C.M = 16 + 6√7~

▪︎$16 + 6√7 × \dfrac{16 + 6√7}{16 + 6√7} + \dfrac{4}{16 + 6√7} × \dfrac{1}{1}$

▪︎$\dfrac{( 16 + 6√7 )²}{16 + 6√7} + \dfrac{4}{16 + 6√7}$

▪︎$\dfrac{( 16 + 6√7 )² + 4}{16 + 6√7}$

☆ According to Formula Used~

▪︎$\dfrac{16² + ( 6√7 )² + 2 × 16 × 6√7}{16 + 6√7}$

▪︎$\dfrac{256 + ( 6² × 7 ) + 192√7}{16 + 6√7}$

▪︎$\dfrac{256 + ( 36 × 7 ) + 192√7}{16 + 6√7}$

▪︎$\dfrac{256 + 252 + 192√7}{16 + 6√7}$

☆ Mixed fraction~

▪︎$13 \ \dfrac{4}{13}$

$\huge\pink{Value \ \ of \ \ 4x² + \dfrac{1}{x²} = \dfrac{173}{13} \ \ or \ \ 13 \ \dfrac{4}{13}}$

Answer:-

Hence, the value of $4x² + \dfrac{1}{x²} = \dfrac{173}{13} \ \ or \ \ 13 \ \dfrac{4}{13}$.

:)