Question

solve for x, using quadratic formula: x-2rootx-6=0​

Answer 1

see the attached document

Answer 2

$\large\underline{\sf{Solution-}}$

The given equation is

$\rm :\longmapsto\:x - 2 \sqrt{x} - 6 = 0$

can be rewritten as

$\rm :\longmapsto\: {( \sqrt{x} )}^{2} - 2 \sqrt{x} - 6 = 0$

Now,

$\rm :\longmapsto\:Its \: a \: quadratic \: in \: \sqrt{x}$

We know,

Quadratic formula

$\rm :\longmapsto\:If \: equation \: is \: {ax}^{2} + bx + c = 0 \: then \:$

$\rm :\longmapsto\:x = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

Here,

$\red{\rm :\longmapsto\:a = 1} \\ \red{\rm :\longmapsto\:b = - 2} \\ \red{\rm :\longmapsto\:c = - 6} \\ \red{\rm :\longmapsto\:x = \sqrt{x}}$

So, on substituting the values we get

$\rm :\longmapsto\: \sqrt{x} = \dfrac{ - ( - 2) \: \pm \: \sqrt{ {( - 2)}^{2} - 4(1)( - 6)} }{2 \times 1}$

$\rm :\longmapsto\: \sqrt{x} = \dfrac{2 \: \pm \: \sqrt{4 + 24} }{2}$

$\rm :\longmapsto\: \sqrt{x} = \dfrac{2 \: \pm \: \sqrt{28} }{2}$

$\rm :\longmapsto\: \sqrt{x} = \dfrac{2 \: \pm \: 2 \sqrt{7} }{2}$

$\rm :\longmapsto\: \sqrt{x} = \dfrac{2(1 \: \pm \: \sqrt{7} )}{2}$

$\rm :\longmapsto\: \sqrt{x} = 1 + \sqrt{7} \: \: or \: \: 1 - \sqrt{7}$

On squaring both sides, we get

$\rm :\longmapsto\:x = 1 + 7 + 2 \sqrt{7} \: \: \: or \: \: \: 1 + 7 - 2 \sqrt{7}$

$\rm :\longmapsto\:x = 8 + 2 \sqrt{7} \: \: \: or \: \: \: 8 - 2 \sqrt{7}$

Aliter method :-

Given equation is

$\rm :\longmapsto\:x - 2 \sqrt{x} - 6 = 0$

$\rm :\longmapsto\:x - 6 = 2 \sqrt{x}$

On squaring both sides, we get

$\rm :\longmapsto\:(x - 6)^{2} = (2 \sqrt{x})^{2}$

$\rm :\longmapsto\: {x}^{2} + 36 - 12x = 4x$

$\rm :\longmapsto\: {x}^{2} + 36 - 12x - 4x = 0$

$\rm :\longmapsto\: {x}^{2} + 36 - 16x = 0$

$\rm :\longmapsto\: {x}^{2} - 16x + 36= 0$

We know,

Quadratic Formula

$\rm :\longmapsto\:If \: equation \: is \: {ax}^{2} + bx + c = 0 \: then \:$

$\rm :\longmapsto\:x = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

Here,

$\red{\rm :\longmapsto\:a = 1} \\ \red{\rm :\longmapsto\:b = - 16} \\ \red{\rm :\longmapsto\:c = 36} \\ \red{\rm :\longmapsto\:x = {x}}$

So, on substituting the values, we get

$\rm :\longmapsto\:x = \dfrac{ - ( - 16) \: \pm \: \sqrt{ {( - 16)}^{2} - 4(1)(36)} }{2 \times 1}$

$\rm :\longmapsto\:x = \dfrac{16 \: \pm \: \sqrt{256 - 144} }{2}$

$\rm :\longmapsto\:x = \dfrac{16 \: \pm \: \sqrt{112} }{2}$

$\rm :\longmapsto\:x = \dfrac{16 \: \pm \: \sqrt{(4)(4)(7)} }{2}$

$\rm :\longmapsto\:x = \dfrac{16 \: \pm \: 4\sqrt{7} }{2}$

$\rm :\longmapsto\:x = \dfrac{2 \: (8 \: \pm \: 2\sqrt{7} )}{2}$

$\rm :\longmapsto\:x = 8 \: \pm \: 2 \sqrt{7}$

$\rm :\longmapsto\:x = 8 + 2 \sqrt{7} \: \: \: or \: \: \: 8 - 2 \sqrt{7}$