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# Solve the following equation:$8 + 9\sqrt {(3x - 1)(x - 2)} = 3{x^2} - 7x$ Verified
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Hint: To make such an equation simple, multiply the expressions written inside the square root and do some operations to make the expression written inside the square root and the right hand side the same.

As, the given equation is
$8 + 9\sqrt {(3x - 1)(x - 2)} = 3{x^2} - 7x$ ………. (1)
First, we multiply the expressions written inside square root of equation 1
$8 + 9\sqrt {3{x^2} - 7x + 2} = 3{x^2} - 7x$ ………… (2)
As, we can see that the expression inside the square root will be the same as the expression on the other side of the equation, if we make a few adjustments.
In this case, we had to add 2 to both sides to get the required equation. So, adding 2 both side in the equation (2), we get
$10 + 9\sqrt {3{x^2} - 7x + 2} = 3{x^2} - 7x + 2$
Now, we can see that there are common terms on both sides of the equation so to simplify the equation, we take the common terms as a variable.
Let $3{x^2} - 7x + 2 = {t^2}$ …………… (3)
Therefore, our equation reduces to
$10 + 9t = {t^2}$ ………….. (4)
Now, equation 4 can be easily solved as it reduces to a quadratic equation.
Solving equation 4 by factorisation,
$\Rightarrow {t^2} - 9t - 10 = 0 \\ \Rightarrow {t^2} - 10t + t - 10 = 0 \\ \Rightarrow t(t - 10) + 1(t - 10) = 0 \\ \Rightarrow (t + 1)(t - 10) = 0 \\ \Rightarrow t = - 1,10 \\$
We get two values of t, so from equation 3,
$\Rightarrow 3{x^2} - 7x + 2 = {t^2}$
When $t = - 1$, then
$\Rightarrow 3{x^2} - 7x + 2 = {( - 1)^2}$
Solving the equation, we get
$\Rightarrow 3{x^2} - 7x + 1 = 0$
$\Rightarrow x = \dfrac{{7 \pm \sqrt {49 - 4(3)(1)} }}{{2(3)}} \\ \Rightarrow x = \dfrac{{7 \pm \sqrt {37} }}{6} \\$
Also, when $t = 10$;
$\Rightarrow 3{x^2} - 7x + 2 = {(10)^2}$
$\Rightarrow 3{x^2} - 7x - 98 = 0 \\ \Rightarrow 3{x^2} - 21x + 14x - 98 = 0 \\ \Rightarrow 3x(x - 7) + 14(x - 7) = 0 \\ \Rightarrow (3x + 14)(x - 7) = 0 \\ \Rightarrow x = - \dfrac{{14}}{3},7 \\$
Hence, the values of x are $\dfrac{{7 - \sqrt {37} }}{6},\dfrac{{7 + \sqrt {37} }}{6}, - \dfrac{{14}}{3}$ and $7$.