Question : If $\sqrt{x}=\sqrt{3}-\sqrt{5}$, then the value of $x^2-16x+6$ is:
Option 1: $0$
Option 2: $-2$
Option 3: $2$
Option 4: $4$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $2$
Solution : Given: $\sqrt{x}=\sqrt{3}-\sqrt{5}$ By squaring both the sides, we get, $x=3+5-2\sqrt{15}$ ⇒ $x-8=-2\sqrt{15}$ Again, squaring both the sides, we get, $x^2-16x+64=60$ ⇒ $x^2-16x+4=0$ $\therefore x^2-16x+6=2$ Hence, the correct answer is $2$.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : If $x^{2} + \frac{1}{x^{2}} = 18$ and $x > 0$, what is the value of $x^{3} + \frac{1}{x^{3}}?$
Question : If $\left(x^2+\frac{1}{x^2}\right)=7$, and $0<x<1$, find the value of $x^2-\frac{1}{x^2}$.
Question : If $x^2-7 x+1=0$ and $0<x<1$, what is the value of $x^2-\frac{1}{x^2}?$
Question : If $x=1+\sqrt{2}+\sqrt{3}$, then the value of $(2x^4-8x^3-5x^2+26x-28)$ is:
Question : If $x>0$ and $x^4+\frac{1}{x^4}=254$, what is the value of $x^5+\frac{1}{x^5}?$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile