Question

Solve $\left( x-5 \right)\left( x-7 \right)\left( x+4 \right)\left( x+6 \right)=504$.

Answer 1

Hint: Find the value of x; take the value of $\left( {{x}^{2}}-x \right)$as t.
Find the roots of t and the substitute it in $t={{x}^{2}}-x$.
Solve the quadratic equations formed and you will get 4 values of x.

Complete step-by-step answer:
Given $\left( x-5 \right)\left( x-7 \right)\left( x+4 \right)\left( x+6 \right)=504$
We can consider $\left[ \left( x-5 \right)\left( x+4 \right) \right]$together and $\left[ \left( x-7 \right)\left( x+6 \right) \right]$.
Open the bracket and form a quadratic equation of the form$a{{x}^{2}}+bx+c$.

& \left[ \left( x-5 \right)\left( x+4 \right) \right]\left[ \left( x-7 \right)\left( x+6 \right) \right]=504-(1) \\\ & \left( x-5 \right)\left( x+4 \right)={{x}^{2}}-5x+4x-20={{x}^{2}}-x-20 \\\ & \left( x-7 \right)\left( x+6 \right)={{x}^{2}}-7x-6x-42={{x}^{2}}-x-42 \\\ \end{aligned}$$ Put $$t={{x}^{2}}-x-(2)$$ $$\begin{aligned} & \Rightarrow \left( t-20 \right)\left( t-42 \right)=504 \\\ & \Rightarrow {{t}^{2}}-20t-42t+840-504=0 \\\ & \Rightarrow {{t}^{2}}-62t+336=0-(3) \\\ \end{aligned}$$ By substituting value of t in equation (1), we can simplify it to equation (3) Now we get a quadratic equation $${{t}^{2}}-62t+336=0$$. The general form $$a{{x}^{2}}+bx+c=0$$, by comparing the general form and equation (3), we get a=1, b=-62 and c=336 substituting these values in $$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$, quadratic formula we get the roots. $$\begin{aligned} & =\dfrac{-\left( -62 \right)\pm \sqrt{{{\left( -62 \right)}^{2}}-4\times 1\times 336}}{2}=\dfrac{62\pm \sqrt{3844-1344}}{2} \\\ & =\dfrac{62\pm \sqrt{2500}}{2}=\dfrac{62\pm 50}{2} \\\ \end{aligned}$$ The roots are $$\left( \dfrac{62+50}{2} \right)$$and $$\left( \dfrac{62-50}{2} \right)$$= 56 and 6 $$\therefore $$Roots of t = 56 and 6 We know, $$t={{x}^{2}}-x$$. Put the values of t = 56. $$\Rightarrow {{x}^{2}}-x-56=0-(4)$$ Now find the roots of equation (4) by using quadratic equation a=1, b = -1, c = -56 $$\begin{aligned} & =\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 1\times \left( -56 \right)}}{2}=\dfrac{-1\pm \sqrt{1+224}}{2} \\\ & =\dfrac{1\pm \sqrt{225}}{2}=\dfrac{1\pm 15}{2} \\\ \end{aligned}$$ The roots are $$\left( \dfrac{1+15}{2} \right)$$and $$\left( \dfrac{1-15}{2} \right)$$= 8 and -7 Similarly, $$t={{x}^{2}}-x$$ , put value of t = 6 $$\Rightarrow {{x}^{2}}-x-6=0$$ a = 1, b = -1, c = -6 $$\begin{aligned} & =\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 1\times \left( -6 \right)}}{2}=\dfrac{-1\pm \sqrt{1+24}}{2} \\\ & =\dfrac{1\pm \sqrt{25}}{2}=\dfrac{1\pm 5}{2} \\\ \end{aligned}$$ The roots are $$\left( \dfrac{1+5}{2} \right)$$and $$\left( \dfrac{1-5}{2} \right)$$= 3 and -2. $$\therefore $$The values of x are 8, -7, 3 and -2. Note: The pair to be multiplied should be chosen in a way that $$t={{x}^{2}}-x$$. Taking $$\left( x-5 \right)\left( x-7 \right)$$and $$\left( x+4 \right)\left( x+6 \right)$$won’t give the required answer. Therefore, we choose $$\left( x-5 \right)\left( x+4 \right)$$and $$\left( x-7 \right)\left( x+6 \right)$$, while forming the equation to get the value of x. Solving the value of t to get the roots.