Question

Operation on real numbers & Alalgebraic identities

If $x = \frac{3 - \sqrt{13}}{2}$ , what is the value of x² + 1/x² ?​

Answer 1

Answer:

Computing x2 and its reciprocal and adding is one way of solving this question (we will do it in method 2).

However, many of these questions can be solved efficiently if we rewrite x2+1x2x2+1x2 in terms of (x−1x)2(x−1x)2or (x+1x)2(x+1x)2.

Which one to choose (x−1x)2(x−1x)2 or (x+1x)2(x+1x)2?

That depends on how the numbers pan out. The first few steps are the same for both. Try which one of the two leads to a more friendly number to calculate and proceed with that.

In this method we will express x2+1x2x2+1x2 in terms of (x−1x)2(x−1x)2

x2+1x2x2+1x2 = (x−1x)2+2(x−1x)2+2

(x−1x)(x−1x) = 3−13√23−132 - 23−13√23−13

= (3−13√)2−42(3−13

Answer 2

• Here we are asked to find the value of

$\sf\red{{x}^{2} + \dfrac{1} {x}^{2}}$and according to the question given that :-

• $\sf \green {\: x = \dfrac{3 - \sqrt{13} }{2}}$

For getting the proper solution, we have to use 2 identities!

• $\sf \purple{ (a + b)² - 2ab = a² + b²}$

• $\sf \purple{ (a - b)² = a² + b² - 2ab}$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀

Here's our required solution!

$:\implies \sf \red{ {x}^{2} + \dfrac{1}{ {x}^{2} } = \bigg( x + \dfrac{1}{x} \bigg) ^{2} - 2 \times x \times \dfrac{1}{x}}$

$: \implies \sf {x}^{2} + \frac{1}{ {x}^{2} }= \bigg( \dfrac{3 - \sqrt{13} }{2} + \frac{1}{ \dfrac{3 - \sqrt{13} }{2} } \bigg) ^{2} - 2$

$:\implies \sf {x}^{2} + \frac{1}{ {x}^{2} } = \bigg( \frac{3 - \sqrt{13} }{2} + \frac{2}{3 - \sqrt{13} } \bigg) ^{2} - 2$

$: \implies \sf {x}^{2} + \frac{1}{ {x}^{2} } = \bigg( \frac{(3 - \sqrt{13} ) ^{2} + 4}{2(3 - \sqrt{13} )} \bigg)^{2} - 2$

$:\implies \sf {x}^{2} + \dfrac{1}{ {x}^{2} } = \bigg( \dfrac{ {3}^{2} + ( \sqrt{13}) ^{2} - 2(3)( \sqrt{13} ) + 4 }{2(3 - \sqrt{13} )} \bigg) ^{2} - 2$

$: \implies \sf \: {x}^{2} + \frac{1}{ {x}^{2} } =\bigg( \frac{9 + 13 - 6 \sqrt{13} + 4}{2(3 - \sqrt{13} )} \bigg) ^{2} - 2$

$: \implies \sf \: {x}^{2} + \frac{1}{ {x}^{2} } = \bigg( \frac{2(13 - 3 \sqrt{13} )}{2(3 - \sqrt{13}) } \bigg) ^{2} - 2$

$: \implies \sf \: {x}^{2} + \frac{1}{ {x}^{2} } = \bigg( \frac{ - \sqrt{13} (3 - \sqrt{13} )}{ 3 - \sqrt{13} } \bigg) ^{2} - 2$

$: \implies \sf {x}^{2} + \frac{1}{ {x}^{2} }= \: {( - \sqrt{13} )}^{2} - 2$

$\sf: \implies {x}^{2} + \frac{1}{ {x}^{2} }= 13 - 2$

$:\implies\underbrace \red {{ \sf {x}^{2} + \frac{1}{ {x}^{2} } = 11}}$