Answer

Verified

450k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

How many crores make 10 million class 7 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths