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$x\left( {2x + 1} \right)\left( {x - 2} \right)\left( {2x - 3} \right) = 63$

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Hint: For solving such types of questions, proceed with multiplying the terms in order to find some common terms and make a substitution with some common term.

Given equation is $x\left( {2x + 1} \right)\left( {x - 2} \right)\left( {2x - 3} \right) = 63$

We will try to separate some common term after multiplying the terms

$

\Rightarrow x\left( {2x + 1} \right)\left( {x - 2} \right)\left( {2x - 3} \right) = 63 \\

\Rightarrow \left\{ {x\left( {2x - 3} \right)} \right\}\left\{ {\left( {2x + 1} \right)\left( {x - 2} \right)} \right\} = 63 \\

\Rightarrow \left\{ {2{x^2} - 3x} \right\}\left\{ {2{x^2} - 4x + x - 2} \right\} = 63 \\

\Rightarrow \left\{ {2{x^2} - 3x} \right\}\left\{ {2{x^2} - 3x - 2} \right\} = 63 \\

$

Now let us substitute some common term

Let $2{x^2} - 3x = t$ -----(1)

So now the equation becomes

$ \Rightarrow t\left( {t - 2} \right) = 63$

Now solving this quadratic equation by simplifying the middle term method

$

\Rightarrow {t^2} - 2t - 63 = 0 \\

\Rightarrow {t^2} - 9t + 7t - 63 = 0 \\

\Rightarrow t\left( {t - 9} \right) + 7\left( {t - 9} \right) = 0 \\

\Rightarrow \left( {t + 7} \right)\left( {t - 9} \right) = 0 \\

$

Hence we get 2 different values of t

$t = - 7,t = 9$

Now putting the value of $t$ back in equation (1), we have

$ \Rightarrow 2{x^2} - 3x = - 7\& 2{x^2} - 3x = 9$

Taking first equation we have

$

\Rightarrow 2{x^2} - 3x = - 7 \\

\Rightarrow 2{x^2} - 3x + 7 = 0 \\

$

We know that roots of any quadratic equation of general form $a{x^2} + bx + c = 0$ are

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

So the roots of first quadratic equation are

$

\Rightarrow x = \dfrac{{ - \left( { - 3} \right) \pm \sqrt {{3^2} - 4\left( 2 \right)\left( 7 \right)} }}{{2\left( 2 \right)}} \\

\Rightarrow x = \dfrac{{3 \pm \sqrt {9 - 56} }}{4} \\

\Rightarrow x = \dfrac{{3 \pm \sqrt { - 47} }}{4} \\

$

Now taking second equation we have

$

\Rightarrow 2{x^2} - 3x = 9 \\

\Rightarrow 2{x^2} - 3x - 9 = 0 \\

\Rightarrow 2{x^2} - 6x + 3x - 9 = 0 \\

\Rightarrow 2x\left( {x - 3} \right) + 3\left( {x - 3} \right) = 0 \\

\Rightarrow \left( {2x + 3} \right)\left( {x - 3} \right) = 0 \\

\Rightarrow x = 3,x = \dfrac{{ - 3}}{2} \\

$

Hence we have 4 roots of the given equation, they are:

$x = \dfrac{{3 + \sqrt { - 47} }}{4},x = \dfrac{{3 - \sqrt { - 47} }}{4},x = 3\& x = \dfrac{{ - 3}}{2}$

Note: Since the quadratic equation is of degree 4 so it must have 4 roots. The problem became easier after substitution of terms in between for simplification. After the substitution the problem of degree 4 reduced to that of degree 2. So simplification is the key to the solution. Also this type of higher degree problem is solved by hit and trial for one root and converting the problem to lower degree.

Given equation is $x\left( {2x + 1} \right)\left( {x - 2} \right)\left( {2x - 3} \right) = 63$

We will try to separate some common term after multiplying the terms

$

\Rightarrow x\left( {2x + 1} \right)\left( {x - 2} \right)\left( {2x - 3} \right) = 63 \\

\Rightarrow \left\{ {x\left( {2x - 3} \right)} \right\}\left\{ {\left( {2x + 1} \right)\left( {x - 2} \right)} \right\} = 63 \\

\Rightarrow \left\{ {2{x^2} - 3x} \right\}\left\{ {2{x^2} - 4x + x - 2} \right\} = 63 \\

\Rightarrow \left\{ {2{x^2} - 3x} \right\}\left\{ {2{x^2} - 3x - 2} \right\} = 63 \\

$

Now let us substitute some common term

Let $2{x^2} - 3x = t$ -----(1)

So now the equation becomes

$ \Rightarrow t\left( {t - 2} \right) = 63$

Now solving this quadratic equation by simplifying the middle term method

$

\Rightarrow {t^2} - 2t - 63 = 0 \\

\Rightarrow {t^2} - 9t + 7t - 63 = 0 \\

\Rightarrow t\left( {t - 9} \right) + 7\left( {t - 9} \right) = 0 \\

\Rightarrow \left( {t + 7} \right)\left( {t - 9} \right) = 0 \\

$

Hence we get 2 different values of t

$t = - 7,t = 9$

Now putting the value of $t$ back in equation (1), we have

$ \Rightarrow 2{x^2} - 3x = - 7\& 2{x^2} - 3x = 9$

Taking first equation we have

$

\Rightarrow 2{x^2} - 3x = - 7 \\

\Rightarrow 2{x^2} - 3x + 7 = 0 \\

$

We know that roots of any quadratic equation of general form $a{x^2} + bx + c = 0$ are

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

So the roots of first quadratic equation are

$

\Rightarrow x = \dfrac{{ - \left( { - 3} \right) \pm \sqrt {{3^2} - 4\left( 2 \right)\left( 7 \right)} }}{{2\left( 2 \right)}} \\

\Rightarrow x = \dfrac{{3 \pm \sqrt {9 - 56} }}{4} \\

\Rightarrow x = \dfrac{{3 \pm \sqrt { - 47} }}{4} \\

$

Now taking second equation we have

$

\Rightarrow 2{x^2} - 3x = 9 \\

\Rightarrow 2{x^2} - 3x - 9 = 0 \\

\Rightarrow 2{x^2} - 6x + 3x - 9 = 0 \\

\Rightarrow 2x\left( {x - 3} \right) + 3\left( {x - 3} \right) = 0 \\

\Rightarrow \left( {2x + 3} \right)\left( {x - 3} \right) = 0 \\

\Rightarrow x = 3,x = \dfrac{{ - 3}}{2} \\

$

Hence we have 4 roots of the given equation, they are:

$x = \dfrac{{3 + \sqrt { - 47} }}{4},x = \dfrac{{3 - \sqrt { - 47} }}{4},x = 3\& x = \dfrac{{ - 3}}{2}$

Note: Since the quadratic equation is of degree 4 so it must have 4 roots. The problem became easier after substitution of terms in between for simplification. After the substitution the problem of degree 4 reduced to that of degree 2. So simplification is the key to the solution. Also this type of higher degree problem is solved by hit and trial for one root and converting the problem to lower degree.

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