Question

please answer it....​

Answer 1

Question :-

Solve the following :-

• $\sf{2x - \dfrac{1}{2x} = 4}$, find : $\sf{4x^2 + \dfrac{1}{4x^2}}$

To find :-

• $\sf{4x^2 + \dfrac{1}{4x^2}}$

Solution :-

• Squaring both sides,

$\sf\bigg({2x - \dfrac{1}{2x}\bigg)^{2} = (4 )^{2} }$

$\sf{(2x)^2 +\bigg (\dfrac{1}{2x}\bigg)^{2} \: - 2 \times (2x) \times \bigg( \dfrac{1}{2x} \bigg) = ( {4})^{2} } \\ \sf{4 {x}^{2} + \dfrac{1}{4x ^{2} } - 2 = 16} \\ \sf{4 {x}^{2} + \dfrac{1}{4x ^{2} } = 16 + 2} \\ \sf{4 {x}^{2} + \dfrac{1}{4x ^{2} } = 18}$

Answer 2

Question :-

• Solve the following :-

• $\sf{2x - \dfrac{1}{2x}} = 4, find : \sf{4x^2 + \dfrac{1}{4x^2}}$

To find :-

• $\sf{4x^2 + \dfrac{1}{4x^2}}$

Solution :-

• Squaring both sides,

$\sf\bigg({2x - \dfrac{1}{2x}\bigg)^{2} = (4 )^{2} }$

$\sf{(2x)^2 +\bigg (\dfrac{1}{2x}\bigg)^{2} \: - 2 \times (2x) \times \bigg( \dfrac{1}{2x} \bigg) = ( {4})^{2} } \\ \sf{4 {x}^{2} + \dfrac{1}{4x ^{2} } - 2 = 16} \\ \sf{4 {x}^{2} + \dfrac{1}{4x ^{2} } = 16 + 2} \\ \sf{4 {x}^{2} + \dfrac{1}{4x ^{2} } = 18}$

Know more

Some useful identities :-

• $\boxed{a + b)^2 = a^2 + b^2 + 2ab}$

• $\boxed{(a - b)^2 = a^2 + b^2 - 2ab}$

• $\boxed{(a + b)^3 = a^3 + b^3 + 3ab(a + b)}$

• $\boxed{(a - b)^3 = a^3 - b^3 - 3ab(a - b)}$