Question

If square root 3 x square - 4 x + 34 + square root 3 x square - 4 x minus 11 is equal to 9 then value of square root 3 x square - 4 x + 34 - square root 3 x square - 4 x minus 11 is

Answer 1

GIVEN :

If $\sqrt{3x^2 - 4 x + 34} +\sqrt{3 x ^2 - 4 x-11}=9$ then the value of $\sqrt{3 x^2 - 4 x + 34} - \sqrt{3 x^2 - 4 x-11}$ is

TO FIND :

The value of $\sqrt{3 x^2 - 4 x + 34} - \sqrt{3 x^2 - 4 x-11}$

SOLUTION :

Given that the expression is $\sqrt{3x^2 - 4 x + 34} +\sqrt{3 x ^2 - 4 x-11}=9$

$\sqrt{3x^2 - 4 x + 34} +\sqrt{3 x ^2 - 4 x-11}=9\hfill (1)$

Let $a =\sqrt{3x^2 -4x + 34}$ and $b =\sqrt{3x^2 - 4x -11}$

Now, (1) becomes,

⇒ a+b = 9

From the algebraic identity :

$(a+b)(a-b)=a^2-b^2$  we have that,

$a-b=\frac{a^2-b^2}{a+b}$

Now substitute the values for a and b we get, $a^2-b^2=(\sqrt{3x^2 -4x + 34})^2-(\sqrt{3x^2 - 4x -11})^2$

$=3x^2-4x+34-(3x^2-4x-11)$

$=3x^2-4x+34-3x^2+4x+11$

= 45

⇒   $a^2-b^2=45$

Substitute the value $a^2-b^2=45$ in $a-b=\frac{a^2-b^2}{a+b}$ we get

$a-b=\frac{45}{9}$

a-b=5

ie., $\sqrt{3 x^2 - 4 x + 34} - \sqrt{3 x^2 - 4 x-11}=5$

∴ the value of $\sqrt{3 x^2 - 4 x + 34} - \sqrt{3 x^2 - 4 x-11}$ is 5