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Solve the following equations:
$\left( 2x-7 \right)\left( {{x}^{2}}-9 \right)\left( 2x+5 \right)=91.$

Last updated date: 29th May 2024
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Hint: Factorize, ${{x}^{2}}-9=\left( x-3 \right)\left( x+3 \right).$ Take average of all four brackets and replace it with another variable. Form biquadratic simpler than given.

Complete step-by-step answer:

The given equation is;
  $\left( 2x-7 \right)\left( {{x}^{2}}-9 \right)\left( 2x+5 \right)=91..........\left( 1 \right)$
As, we know ${{a}^{2}}-{{b}^{2}}=\left( a-6 \right)\left( a+b \right)$; we can use this identity with ${{x}^{2}}-9={{x}^{2}}-{{3}^{2}}$ and replace it with $\left( x-3 \right)\left( x+3 \right)$. Hence, equation (1) will become;
$\left( 2x-7 \right)\left( x-3 \right)\left( x+3 \right)\left( 2x+5 \right)=91$
Taking out ‘2’ as a common from first bracket and last bracket of above equation, we get
  & 2\left( x-\dfrac{7}{2} \right)\left( x-3 \right)\left( x+3 \right)2\left( x+\dfrac{5}{2} \right)=91 \\
 & or \\
 & 4\left( x-\dfrac{7}{2} \right)\left( x-3 \right)\left( x+3 \right)\left( x+\dfrac{5}{2} \right)=91......\left( 2 \right) \\
Now, let us take another variable ‘y’ which can be obtained by getting average of all brackets as follows;
  & y=\dfrac{x-\dfrac{7}{2}+x-3+x+3+x+\dfrac{5}{2}}{4} \\
 & y=\dfrac{4x-1}{4} \\
 & y=x-\dfrac{1}{4} \\
 & or \\
 & x=y+\dfrac{1}{4}........................\left( 3 \right) \\
Now, we can replace variable ‘x’ by ‘y’ in equation (2) with the help of relation obtained in equation (3);
Hence, equation (3) in variable ‘y’ can be expressed as;
  & 4\left( y+\dfrac{1}{4}-\dfrac{7}{2} \right)\left( y+\dfrac{1}{4}-3 \right)\left( y+\dfrac{1}{4}+3 \right)\left( y+\dfrac{1}{4}+\dfrac{5}{2} \right)=91 \\
 & 4\left( y+\dfrac{1-14}{4} \right)\left( y+\dfrac{1-12}{4} \right)\left( y+\dfrac{1+12}{4} \right)\left( y+\dfrac{1+10}{4} \right)=91 \\
 & 4\left( y-\dfrac{13}{4} \right)\left( y-\dfrac{11}{4} \right)\left( y+\dfrac{13}{4} \right)\left( y+\dfrac{11}{4} \right)=91 \\
 & Or \\
 & 4\left( y-\dfrac{13}{4} \right)\left( y+\dfrac{13}{4} \right)\left( y-\dfrac{11}{4} \right)\left( y+\dfrac{11}{4} \right)=91...........\left( 4 \right) \\
As, we know $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}.$ therefore, we can replace
  & \left( y-\dfrac{13}{4} \right)\left( y+\dfrac{13}{4} \right)by\text{ }{{y}^{2}}-{{\left( \dfrac{13}{4} \right)}^{2}}, \\
 & and \\
 & \left( y-\dfrac{11}{4} \right)\left( y+\dfrac{11}{4} \right)by\text{ }{{y}^{2}}-{{\left( \dfrac{11}{4} \right)}^{2}} \\
Hence, rewriting equation (4) , we get
  & 4\left( {{y}^{2}}-{{\left( \dfrac{13}{4} \right)}^{2}} \right)\left( {{y}^{2}}-\dfrac{{{\left( 11 \right)}^{2}}}{{{\left( 4 \right)}^{2}}} \right)=91 \\
 & 4\left( {{y}^{2}}-\dfrac{169}{16} \right)\left( {{y}^{2}}-\dfrac{121}{16} \right)=91 \\
Taking ${{y}^{2}}=z$ to minimize the complexity of problem in above equation, we get
  & 4\left( z-\dfrac{169}{16} \right)\left( z-\dfrac{121}{16} \right)=91 \\
 & 4\dfrac{\left( 16z-169 \right)\left( 16z-121 \right)}{16\times 16}=91 \\
On simplifying the above equation and cross multiplying as well, we get;
$\left( 16z-169 \right)\left( 16z-121 \right)=64\times 91$
Multiplying both the brackets, we get
  & 256{{z}^{2}}-121\times 16z-169\times 16z+169\times 121=64\times 91 \\
 & 256{{z}^{2}}-290z\times 16+169\times 121=64\times 91 \\
 & 256{{z}^{2}}-4640z+14625=0.............\left( 5 \right) \\
Now, we have a quadratic equation, so it will have two roots and can be given by quadratic formula as given below;
If, we have any quadratic equation,$A{{x}^{2}}Bx+C=0$, then roots of quadratic equation are;
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$
Now, using the quadratic formula with equation (5), we get;
  & z=\dfrac{4640\pm \sqrt{{{\left( 4640 \right)}^{2}}-4\times 256\times 14625}}{2\times 256} \\
 & z=\dfrac{4640\pm \sqrt{6553600}}{2\times 256} \\
Now, as we know that $\sqrt{6553600}=2560$ ,
Hence, roots can be simplified as;
  & z=\dfrac{4640\pm 2560}{512} \\
 & z=\dfrac{4640}{512}\pm \dfrac{2560}{512} \\
Simplifying z, we get;
$z=\dfrac{145}{16}\pm 5$
Now, we have two roots as
  & z=\dfrac{145+80}{16} \\
 & and \\
 & z=\dfrac{145-80}{16} \\
 & z=\dfrac{225}{16},\dfrac{65}{16}......................\left( 6 \right) \\
Now, we have ${{y}^{2}}=z$
  & y=\pm \sqrt{\dfrac{225}{16}},y=\pm \sqrt{\dfrac{65}{16}} \\
 & Or \\
 & y=\dfrac{\pm 15}{4},y=\dfrac{\pm \sqrt{65}}{4} \\
Now, for calculating x, we have to use equation (3) i.e. \[x=y+\dfrac{1}{4}\];
Hence, we get
  & x=\dfrac{\pm 15}{4}+\dfrac{1}{4} \\
 & and \\
 & x=\pm \dfrac{\sqrt{65}}{4}+\dfrac{1}{4} \\
Now four values of x, can be given as
  & x=\dfrac{15+1}{4},\dfrac{-15+1}{4},\dfrac{\sqrt{65}+1}{4},\dfrac{-\sqrt{65}+1}{4} \\
 & Or \\
 & x=4,\dfrac{-7}{2},\dfrac{\sqrt{65}+1}{4},\dfrac{-\sqrt{65}+1}{4} \\
Therefore, values of ‘x’ after solving the given equation are;
$\left( 4,\dfrac{-7}{2},\dfrac{\sqrt{65}+1}{4},\dfrac{-\sqrt{65}+1}{4} \right)$

Note: One can multiply the brackets to get bi-quadratic equation but it is really complex to get all four solutions from that equation. We have to put a lot of value into getting roots. And one cannot predict $\dfrac{-7}{2}$ as a root. Hence, this approach is more complex than the given solution.
Calculation is an important task of this question as we have to multiply bigger numbers and calculating square root of 6553600, one can go wrong with calculations as well.
Taking an average of four brackets is the key point of the question.