Solve the following equation:
$8 + 9\sqrt {(3x - 1)(x - 2)} = 3{x^2} - 7x$
Answer
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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.
Complete step-by-step answer:
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$
Solving the above quadratic equation using the quadratic formula
$
\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}$
Solving the equation using factorisation, we get
$
\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 $7$.
Note: In such types of problems, we have to simplify the equation in such a way that the whole equation can be represented by another variable but in a simpler form which is easily solvable.
Complete step-by-step answer:
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$
Solving the above quadratic equation using the quadratic formula
$
\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}$
Solving the equation using factorisation, we get
$
\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 $7$.
Note: In such types of problems, we have to simplify the equation in such a way that the whole equation can be represented by another variable but in a simpler form which is easily solvable.
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