Question

Plz answer the question ......

Answer 1

Let,

$\displaystyle\longrightarrow\sf{3x^2-4x=k}$

Then the given equation becomes,

$\displaystyle\longrightarrow\sf{k+\sqrt{k-6}=18}$

$\displaystyle\longrightarrow\sf{\sqrt{k-6}=18-k}$

$\displaystyle\longrightarrow\sf{k-6=324-36k+k^2}$

$\displaystyle\longrightarrow\sf{k^2-37k+330=0}$

$\displaystyle\longrightarrow\sf{k^2-15k-22k+330=0}$

$\displaystyle\longrightarrow\sf{k(k-15)-22(k-15)=0}$

$\displaystyle\longrightarrow\sf{(k-15)(k-22)=0}$

From this we have two cases.

Case 1:-

$\displaystyle\longrightarrow\sf{k-15=0}$

$\displaystyle\longrightarrow\sf{3x^2-4x-15=0}$

$\displaystyle\longrightarrow\sf{3x^2-9x+5x-15=0}$

$\displaystyle\longrightarrow\sf{3x(x-3)+5(x-3)=0}$

$\displaystyle\longrightarrow\sf{(x-3)(3x+5)=0}$

$\displaystyle\Longrightarrow\sf{x=3\quad\quad;\quad\quad x=-\dfrac{5}{3}}$

Case 2:-

$\displaystyle\longrightarrow\sf{k-22=0}$

$\displaystyle\longrightarrow\sf{3x^2-4x-22=0}$

$\displaystyle\longrightarrow\sf{x=\dfrac{4\pm\sqrt{4^2+4\times3\times22}}{6}}$

$\displaystyle\longrightarrow\sf{x=\dfrac{4\pm\sqrt{280}}{6}}$

$\displaystyle\longrightarrow\sf{x=\dfrac{2+\sqrt{70}}{3}\quad\quad;\quad\quad x=\dfrac{2-\sqrt{70}}{3}}$

So possible values of x got are,

$\displaystyle\longrightarrow\sf{\underline{\underline{x=3\quad;\quad x=-\dfrac{5}{3}\quad;\quad x=\dfrac{2+\sqrt{70}}{3}\quad;\quad x=\dfrac{2-\sqrt{70}}{3}}}}$

But before we should also check that, for which values of x,

$\displaystyle\longrightarrow\sf{3x^2-4x-6\geq0}$

But on checking the four values of x are possible.