If 4 x^{4}=5 x^{2}-1, x > \frac{1}\sqrt{2}, then what is the value of (2 x^{2}-\sqrt{2}-1) ?

SSC CGL 17 Aug 2021 Shift 3 Paper

Question : 57 of 100

Marks: +1, -0

4 x^{4}=5 x^{2}-1, x > \;{1}/{√{2}}

, then what is the value of

(2 x^{2}-√{2}-1) ?

Solution:

Given,

4 x^{4}=5 x^{2}-1

⇒ \;\; 4 x^{4}-5 x^{2}+1=0

⇒ \;\; 4 x^{4}-4 x^{2}-x^{2}+1=0

⇒ \;\; 4 x^{2}(x^{2}-1)-1(x^{2}-1)=0

⇒ \;\; (4 x^{2}-1)(x^{2}-1)=0

⇒ \;\; x^{2}=\;{1}/{4}

and

x^{2}=1

⇒ \;\; x=\;{1}/{2}

and

x=1

We can't take

x=\;{1}/{2}

because the value of

x

is more then

\;{1}/{√{2}}

.
Now, we will take

x=1

∴ 2 x^{2}-1-x=2(1)^{2}-1-1=2-2=0

Go to Question:

Prev Question

Next Question