Question

if x - 1/x = 3. find the value of x²+ 1/x² and x⁴ + 1/x⁴ (i)4,88 (ii)7,100 (iii)10,112 (iv)11,119​

Answer 1

Answer:

iv) 11,119

Step-by-step explanation:

x-1/x=3

squareing on both side

(x-1/x)²=3²

x²+1/x²-2*1/x*x=9

x²+1/x²=9+2

=11

x²+1/x²=11

squareing on both sides

(x²+1/x²)²=11²

x⁴+1/x⁴+2=121

x⁴+1/x⁴=119

Answer 2

★GIVEN★

$\large{\sf{x-\dfrac{1}{x}=3}}$

★To Find★

• $\large{\sf{{x}^{2}+\dfrac{1}{{x}^{2}}}}$

• $\large{\sf{{x}^{4}+\dfrac{1}{{x}^{4}}}}$

★SOLUTION★

$\large{\green{\underline{\underline{\sf{1))\:{x}^{2}+\dfrac{1}{{x}^{2}}}}}}}$

By using the identify,

$\large{\green{\boxed{\bf{{(a-b)}^{2}={a}^{2}-2ab+{b}^{2}}}}}$

According to the question,

Squaring both sides,

$\large\implies{\sf{{(x-\dfrac{1}{x}})^{2}={x}^{2}-2.x.\dfrac{1}{x}+\dfrac{1}{{x}^{2}}}}$

$\large\implies{\sf{{3}^{2}={x}^{2}-2.\:\cancel{x}\:.\dfrac{1}{\cancel{x}}+\dfrac{1}{{x}^{2}}}}$

$\large\implies{\sf{9={x}^{2}-2+\dfrac{1}{{x}^{2}}}}$

$\large\implies{\sf{9+2={x}^{2}+\dfrac{1}{{x}^{2}}}}$

$\large\implies{\sf{11={x}^{2}+\dfrac{1}{{x}^{2}}}}$

$\large\therefore\boxed{\bf{{x}^{2}+\dfrac{1}{{x}^{2}}=11}}$

$\large{\green{\underline{\underline{\sf{2))\:{x}^{4}+\dfrac{1}{{x}^{4}}}}}}}$

By using the identify,

$\large{\green{\boxed{\bf{{(a+b)}^{2}={a}^{2}+2ab+{b}^{2}}}}}$

According to the question,

Squaring both the sides,

$\large\implies{\sf{({{x}^{2}+\dfrac{1}{{x}^{2}}})^{2}={x}^{4}+2.{x}^{2}.\dfrac{1}{{x}^{2}}+\dfrac{1}{{x}^{4}}}}$

$\large\implies{\sf{{11}^{2}={x}^{4}+2.\:\cancel{{x}^{2}}\:.\dfrac{1}{\cancel{{x}^{2}}}+\dfrac{1}{{x}^{4}}}}$

$\large\implies{\sf{121={x}^{4}+2+\dfrac{1}{{x}^{4}}}}$

$\large\implies{\sf{121-2={x}^{4}+\dfrac{1}{{x}^{4}}}}$

$\large\implies{\sf{119={x}^{4}+\dfrac{1}{{x}^{4}}}}$

$\large\therefore\boxed{\bf{{x}^{4}+\dfrac{1}{{x}^{4}}=119}}$

So your answer is option (iv) 11,119.