Solve the following equations: $\left( x-7 \right)\left( x-3 \right)\left( x+5 \right)\left( x+1 \right)=1680$.
Last updated date: 24th Mar 2023
•
Total views: 306.3k
•
Views today: 6.83k
Answer
306.3k+ views
Hint: Take the average of all brackets given in the LHS of the equation and take it as another variable and form a new biquadratic equation. Now, solve it further to get values of that variable and hence values of ‘x’ as well.
Complete step-by-step answer:
The given equation is,
$\left( x-7 \right)\left( x-3 \right)\left( x+5 \right)\left( x+1 \right)=1680$…………….. (i)
Let us change the given variables ‘x’ by taking average of all brackets as a variables ’y’ in following way:-
$\Rightarrow$ $y=\dfrac{\left( x-7 \right)+\left( x-3 \right)+\left( x+5 \right)+\left( x+1 \right)}{4}$
$\Rightarrow$ $y=\dfrac{4x-4}{4}=x-1$
Hence, we get the value ’y’ as ‘x-1’. So, now ‘x’ can be written in form ‘y’ in following way:-
x=y+1………………(ii)
Let us replace variable ‘x’ from equation (i) with the help of equation (ii) hence we get,
$\Rightarrow$ $\left( y+1-7 \right)\left( y+1-3 \right)\left( y+1+5 \right)\left( y+1+1 \right)=1680$
$\Rightarrow$ \[\left( y-6 \right)\left( y-2 \right)\left( y+6 \right)\left( y+2 \right)=1680\]
Rewriting the above equation we get,
$\Rightarrow$ $\left( y-6 \right)\left( y+6 \right)\left( y-2 \right)\left( y+2 \right)=1680$…………. (iii)
Now, we can use an algebraic identity given as
$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
So, we can simplify the equation (iii) using the above relation, we get
$\Rightarrow$ $\left( {{y}^{2}}-36 \right)\left( {{y}^{2}}-4 \right)=1680$……………. (iv)
Let ${{y}^{2}}=t$, so that we can solve the above equation in a simpler way. Hence, we get
$\Rightarrow$ $\left( t-36 \right)\left( t-4 \right)=1680$
Now, multiplying the both brackets, we get
$\begin{align}
& {{t}^{2}}-4t-36t+36\times 4=1680 \\
& {{t}^{2}}-40t+144=1680 \\
\end{align}$
${{t}^{2}}-40t-1536=0$………. (v)
Now we can get values of ‘t’ from the above equation by using quadratic formula. We know roots of quadratic $A{{x}^{2}}+Bx+C=0$ can be given as
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$
Hence, value of ‘t’ from equation (V) be
$\begin{align}
& t=\dfrac{-(-40)\pm \sqrt{{{(-40)}^{2}}-4\times 1\times (-1536)}}{2\times 1} \\
& t=\dfrac{40\pm \sqrt{1600+4\times 1536}}{2} \\
& t=\dfrac{40\pm \sqrt{7744}}{2} \\
\end{align}$
Now, we can calculate the square root of 7744. So, we get
$t=\dfrac{40\pm 88}{2}$
So, two values of ‘t’ can be given as
$t=\dfrac{40+88}{2}$or$t=\dfrac{40-88}{2}$
Or
t=64 or t=-24
We had assumed $t={{y}^{2}}$; hence, we get
As we know squares of any number cannot be negative. ${{y}^{2}}$can never be -24.
So, we can ignore it. Hence,
${{y}^{2}}=64$
Taking square root on both sides, we get
$y=\pm 8$……………(vi)
Now, we can use relation x=y+1 from equation (ii) to get values of ‘x’. Hence,
x =8+1 and x=-8+1
x =9 and x=-7
So, values of ’x’ satisfying the given equation are 9, -7.
Note: One can think that a given polynomial is of degree’4’ so why is there only two roots 9 and -7. We can get two values of x from the relation ${{x}^{2}}=-24$which will be imaginary and will be in terms ‘i’. So imaginary roots can be given as
$x=2\sqrt{6i}$and$x=-2\sqrt{6i}$
Where,
\[i=\sqrt{-1}\]
So, there are two real and two imaginary roots of the given equation.
One can multiply the given brackets in the equation and can form a 4 degree polynomial. But guessing the roots 9 and -7 is very complex and difficult. So, take the average of all brackets and replace the given variable with another in these kinds of questions.
Complete step-by-step answer:
The given equation is,
$\left( x-7 \right)\left( x-3 \right)\left( x+5 \right)\left( x+1 \right)=1680$…………….. (i)
Let us change the given variables ‘x’ by taking average of all brackets as a variables ’y’ in following way:-
$\Rightarrow$ $y=\dfrac{\left( x-7 \right)+\left( x-3 \right)+\left( x+5 \right)+\left( x+1 \right)}{4}$
$\Rightarrow$ $y=\dfrac{4x-4}{4}=x-1$
Hence, we get the value ’y’ as ‘x-1’. So, now ‘x’ can be written in form ‘y’ in following way:-
x=y+1………………(ii)
Let us replace variable ‘x’ from equation (i) with the help of equation (ii) hence we get,
$\Rightarrow$ $\left( y+1-7 \right)\left( y+1-3 \right)\left( y+1+5 \right)\left( y+1+1 \right)=1680$
$\Rightarrow$ \[\left( y-6 \right)\left( y-2 \right)\left( y+6 \right)\left( y+2 \right)=1680\]
Rewriting the above equation we get,
$\Rightarrow$ $\left( y-6 \right)\left( y+6 \right)\left( y-2 \right)\left( y+2 \right)=1680$…………. (iii)
Now, we can use an algebraic identity given as
$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
So, we can simplify the equation (iii) using the above relation, we get
$\Rightarrow$ $\left( {{y}^{2}}-36 \right)\left( {{y}^{2}}-4 \right)=1680$……………. (iv)
Let ${{y}^{2}}=t$, so that we can solve the above equation in a simpler way. Hence, we get
$\Rightarrow$ $\left( t-36 \right)\left( t-4 \right)=1680$
Now, multiplying the both brackets, we get
$\begin{align}
& {{t}^{2}}-4t-36t+36\times 4=1680 \\
& {{t}^{2}}-40t+144=1680 \\
\end{align}$
${{t}^{2}}-40t-1536=0$………. (v)
Now we can get values of ‘t’ from the above equation by using quadratic formula. We know roots of quadratic $A{{x}^{2}}+Bx+C=0$ can be given as
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$
Hence, value of ‘t’ from equation (V) be
$\begin{align}
& t=\dfrac{-(-40)\pm \sqrt{{{(-40)}^{2}}-4\times 1\times (-1536)}}{2\times 1} \\
& t=\dfrac{40\pm \sqrt{1600+4\times 1536}}{2} \\
& t=\dfrac{40\pm \sqrt{7744}}{2} \\
\end{align}$
Now, we can calculate the square root of 7744. So, we get
$t=\dfrac{40\pm 88}{2}$
So, two values of ‘t’ can be given as
$t=\dfrac{40+88}{2}$or$t=\dfrac{40-88}{2}$
Or
t=64 or t=-24
We had assumed $t={{y}^{2}}$; hence, we get
As we know squares of any number cannot be negative. ${{y}^{2}}$can never be -24.
So, we can ignore it. Hence,
${{y}^{2}}=64$
Taking square root on both sides, we get
$y=\pm 8$……………(vi)
Now, we can use relation x=y+1 from equation (ii) to get values of ‘x’. Hence,
x =8+1 and x=-8+1
x =9 and x=-7
So, values of ’x’ satisfying the given equation are 9, -7.
Note: One can think that a given polynomial is of degree’4’ so why is there only two roots 9 and -7. We can get two values of x from the relation ${{x}^{2}}=-24$which will be imaginary and will be in terms ‘i’. So imaginary roots can be given as
$x=2\sqrt{6i}$and$x=-2\sqrt{6i}$
Where,
\[i=\sqrt{-1}\]
So, there are two real and two imaginary roots of the given equation.
One can multiply the given brackets in the equation and can form a 4 degree polynomial. But guessing the roots 9 and -7 is very complex and difficult. So, take the average of all brackets and replace the given variable with another in these kinds of questions.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

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

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
