Rajasthan BoardEnglish MediumSTD 9MATHSAlgebraic Identities4 Marks
Question
If $\text{x}-\frac{1}{\text{x}}=\frac{1}{2},$ then write the value of $4\text{x}^2+\frac{4}{\text{x}^2}.$
✓
Answer
We have to find the value of $4\text{x}^2+\frac{4}{\text{x}^2}$
Given $\text{x}-\frac{1}{\text{x}}=\frac{1}{2}$
Using identity $(a - b)^2 = a^2 - 2ab + b^2$
Here $\text{a} = \text{x},\ \text{b}=\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2-2\times\text{x}\times\frac{1}{\text{x}}+\Big(\frac{1}{\text{x}}\Big)^2$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2-2\times\not\text{x}\times\frac{1}{\not\text{x}}+\frac{1}{\text{x}}\times\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2-2+\frac{1}{\text{x}^2}$
By substituting the value of $\text{x}-\frac{1}{\text{x}}=\frac{1}{2}$ we get
$\Big(\frac{1}{2}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
By transposing -2 to left hand side we get
$\frac{1}{4}+2=\text{x}^2+\frac{1}{\text{x}^2}$
By taking least common multiply we get
$\frac{1}{4}+\frac{2}{1}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{1}{4}+\frac{2}{1}\times\frac{4}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{1}{4}+\frac{8}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{1+8}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{+9}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
By multiplying 4 on both sides we get
$4\times\frac{9}{4}=4\text{x}^2+4\times\frac{1}{\text{x}^2}$
$4\times\frac{9}{4}=4\text{x}^2+\frac{4}{\text{x}^2}$
$9 = 4\text{x}^2+\frac{4}{\text{x}^2}$
Hence the value of $4\text{x}^2+\frac{4}{\text{x}^2}$ is $9.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.