Question

Find the number of real roots of the equation \[{\left( {6 - x} \right)^4} + {\left( {8 - x} \right)^4} = 16\].
A. 0
B. 4
C. 2
D. None of these

Answer 1

Hint: First substitute some other value for \[x\] and simplify the given equation. Then use binomial expansion to expand the terms of the equation. In the end, factorize the simplified equation and find the roots of the equation.

Formula Used: Binomial expansion theorem: \[{\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {{}^n} {C_r}{a^{\left( {n - r} \right)}}{b^r}\]

Complete step by step solution:
The given equation is \[{\left( {6 - x} \right)^4} + {\left( {8 - x} \right)^4} = 16\].
Let’s simplify the given equation.
Substitute \[7 - x = y\] in the above equation.
Then,
\[{\left( {y - 1} \right)^4} + {\left( {y + 1} \right)^4} = 16\]
Now apply binomial expansion theorem.
\[\left( {{y^4} - 4{y^3} + 6{y^2} - 4y + 1} \right) + \left( {{y^4} + 4{y^3} + 6{y^2} + 4y + 1} \right) = 16\]
Solve the above equation.
\[2{y^4} + 12{y^2} + 2 = 16\]
Divide both sides by 2.
\[{y^4} + 6{y^2} + 1 = 8\]
\[ \Rightarrow \]\[{y^4} + 6{y^2} - 7 = 0\]
Now factorize the above equation.
\[{y^4} + 7{y^2} - {y^2} - 7 = 0\]
\[ \Rightarrow \]\[{y^2}\left( {{y^2} + 7} \right) - 1\left( {{y^2} + 7} \right) = 0\]
\[ \Rightarrow \]\[\left( {{y^2} + 7} \right)\left( {{y^2} - 1} \right) = 0\]
\[ \Rightarrow \]\[\left( {{y^2} + 7} \right) = 0\] or \[\left( {{y^2} - 1} \right) = 0\]
Since the roots are real. So \[{y^2} > 0\].
Therefore, the only solution is \[\left( {{y^2} - 1} \right) = 0\].
\[ \Rightarrow \]\[{y^2} = 1\]
Take square root on both sides.
\[y = \pm 1\]

Now resubstitute \[y = \pm 1\] in the equation \[7 - x = y\]. We get,
When \[y = 1\]:
\[x = 7 - 1\]
\[ \Rightarrow \]\[x = 6\]

When \[y = - 1\]:
\[x = 7 - \left( { - 1} \right)\]
\[ \Rightarrow \]\[x = 8\]

Thus, the number of real roots is 2.

Hence the correct option is C.

Note:The root of an equation is a value of a variable or a solution for which the equation is true.
A real root of an equation is a real number.