Question

If [x+1uponx]=12,find the value of [x²+1uponx²​

Answer 1

Qᴜᴇsᴛɪᴏɴ:-If [x+1uponx]=12,find the value of [x²+1uponx²​

Answer:

${x }^{2} + \frac{1}{x^2} = 142$


Solution:

Given,

$x + \frac{1}{x} = 12$


Use the following algebraic identity,

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


Then,

${x}^{2} + \frac{1}{x^2} = (x + \frac{1}{x} )^{2} - 2 \times x \times \frac{1}{x}$
${x}^{2} + \frac{1}{x^2} = (x + \frac{1}{x}) {}^{2} - 2$


Thus found,
Sᴜʙsᴛɪᴛᴜᴅᴇ,
$x + \frac{1}{x} = 12$
${x}^{2} + \frac{1}{x^2} = {12}^{2} - 2$
${x}^{2} + \frac{1}{x^2} = 144 - 2$
${x}^{2} + \frac{1}{x^2} = 142$

Answer 2

Step-by-step explanation:

Uᴘᴀʀ ᴊɪsɴᴇ ᴀɴsᴡᴇʀ ᴅɪʏᴀ ʜᴀɪ ᴜsᴋᴏ $\huge\mathfrak\color{red}{follow}$ karo plzz