If 6x^ 2 - 5x - 3 x^ 2 - 2x + 6 ≤ 4, then the least and highest...

We have,

$\dfrac{6x^{2} - 5x - 3}{x^{2} - 2x + 6} \le 4$

$⇒ \dfrac{6x^{2} - 5x - 3 - 4x^{2} + 8x - 24}{x^{2} - 2x + 6} \le 0$

$⇒ \dfrac{2x^{2} + 3x - 27}{x^{2} - 2x + 6} \le 0$

$⇒ \dfrac{\left(2x + 9\right)\left(x - 3\right)}{\left(x - 1\right)2 + 5} \le 0$

Now,~\left(x - 1\right)2 + 5 > 0, \forall x \in R$.</p><p>So,</p><p>$2x^{2} + 3x - 27 \le 0$</p><p>$⇒ \left(2x + 9\right)\left(x - 3\right) \le 0$</p><p>~$⇒ - \dfrac{9}{2} \le x \le 3$</p><p>Now, at~$x = - \dfrac{9}{2}$,~$4x^{2} = 4 \times \left(\dfrac{- 9}{2}\right)2 = 81$</p><p>And, at~$x = 3$,~$4x^{2} = 36$</p><p>Therefore, greatest value of~$4x^{2}$~is~$81$.</p><p>And, least value of~$4x^{2}$~is$0$~occurs at~$x = 0$.</p></blockquote>$