Question

solve the following √x^2-3x=4x^2-12x-3
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Answer 1

Given:

The equation -  $- \[\sqrt {{x^2}} - 3x = 4{x^2} - 12x - 3\]$

To Find:

Solve for x.

Formula Used:

$\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]$

Solution:

Firstly simplify the given equation-

$\[\begin{gathered} \sqrt {{x^2}} - 3x = 4{x^2} - 12x - 3 \hfill \\ x - 3x = 4{x^2} - 12x - 3 \hfill \\ - 2x = 4{x^2} - 12x - 3 \hfill \\ 4{x^2} - (12 + 2) x - 3 = 0 \hfill \\ 4{x^2} - 10x - 3 = 0 \hfill \\ \end{gathered} \]$

Here, a=4, b= -10 and c= -3

On substituting the value of a, b and c in formula, we get $\[\begin{gathered} x = \dfrac{{ - ( - 10) \pm \sqrt {{{(-10)}^2} - 4 \times 4 \times ( - 3)} }}{{2 \times 4}} \hfill \\ x = \dfrac{{10 \pm \sqrt {100 + 48} }}{8} \hfill \\ x = \dfrac{{10 \pm \sqrt {148} }}{8} \hfill \\ x = \dfrac{{10 \pm 2\sqrt {37} }}{8} \hfill \\ x = \dfrac{{5 \pm \sqrt {37} }}{4} \hfill \\ \end{gathered} \]$

Hence,$\[x = \dfrac{{5 \pm \sqrt {37} }}{4}\]$

Answer 2

$Given \: \sqrt{x^{2} - 3x } = 4x^{2} - 12x - 3$

$\implies \sqrt{x^{2} - 3x } = 4(x^{2} - 3x )- 3$

$\blue { Let \: a = x^{2} - 3x \: --(1) }$

$\implies \sqrt{a} = 4a - 3$

/* On squaring both sides ,we get */

$\implies a = (4a-3)^{2}$

$\implies a = (4a)^{2} - 2 \times (4a) \times 3 + 3^{2}$

$\implies 0 = 16a^{2} - 24a + 9 - a$

$\implies 16a^{2} - 25a + 9 = 0$

/* Splitting the middle term, we get */

$\implies 16a^{2} - 16a - 9a + 9 = 0$

$\implies 16a(a - 1) - 9(a-1) = 0$

$\implies (a-1)(16a-9) = 0$

$\implies a - 1 = 0 \:Or \: 16a - 9 = 0$

$\implies x^{2}-3x-1 = 0\:or \: 16(x^{2}-3x)-9=0$

/* From (1) */

$i) Compare \: x^{2}-3x-1=0 \: with \\ ax^{2}+bx+c=0 , we \:get$

$a = 1 , b = -3 \:and \: c = -1$

$Discreminant (D) \\= b^{2} - 4ac\\= (-3)^{2} -4 \times 1 \times (-1) \\= 9 + 4\\= 13$

Using Quadratic formula :

$x = \frac{-b \pm \sqrt{D}}{2a}$

$= \frac{ -(-3) \pm \sqrt{13}}{2 \times 1 } \\= \frac{3\pm \sqrt{13}}{2}$

Therefore.,

$\red{x} \green { = \frac{3\pm \sqrt{13}}{2} }$

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