Question : If $x^2-3.2 x+1=0$ and $x>1$, the value of $x^2-\frac{1}{x^2}$ is:
Option 1: $16.8 \sqrt{0.39}$
Option 2: $12.8 \sqrt{0.39}$
Option 3: $16.8 \sqrt{0.32}$
Option 4: $12.8 \sqrt{0.32}$
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $12.8 \sqrt{0.39}$
Solution : Given: $x^2-3.2 x+1=0$ and $x>1$ $x^2-3.2 x+1=0$ Dividing both sides by $x$, we get, $⇒x-3.2 +\frac{1}{x}=0$ $⇒x+\frac{1}{x}=3.2 $---(1) Squaring both sides, we get, $⇒x^2+\frac{1}{x^2}+2=10.24 $ $⇒x^2+\frac{1}{x^2}-2=6.24 $ $⇒(x-\frac{1}{x})^2=6.24 $ $⇒(x-\frac{1}{x})=\sqrt{6.24}$---(2) Multiplying 1 and 2, $(x+\frac{1}{x})(x-\frac{1}{x})=3.2×\sqrt{6.24}=3.2×\sqrt{16×0.39}=12.8 \sqrt{0.39}$ Hence, the correct answer is $12.8 \sqrt{0.39}$.
Candidates can download this e-book to give a boost to thier preparation.
Answer Key | Eligibility | Application | Admit Card | Preparation Tips | Result | Cutoff
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile