Determine. if 3 is a root of the equation given below:

Question:

Determine. if 3 is a root of the equation given below:

$\sqrt{x^{2}-4 x+3}+\sqrt{x^{2}-9}=\sqrt{4 x^{2}-14 x+16}$

Solution:

We have been given that,

$\sqrt{x^{2}-4 x+3}+\sqrt{x^{2}-9}=\sqrt{4 x^{2}-14 x+16}$

We have to check whether x = 3 is the solution of the given equation or not.

Now, if $x=3$ is a root of the above quadratic equation, then it should satisfy the whole. So substituting $x=3$ in the above equation, we have,

Left hand side

$=\sqrt{(3)^{2}-4(3)+3}+\sqrt{(3)^{2}-9}$

$=\sqrt{0}+\sqrt{0}$

$=0$

Right hand side

$=\sqrt{4\left(3^{2}\right)-14(3)+16}$

$=\sqrt{36-42+16}$

$=\sqrt{10}$

Now since, we can see from above that left hand side and right hand side are not equal. Therefore $x=3$ is not the solution of the given quadratic equation.

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now