Question

The number of real roots of ${{\left( x+3 \right)}^{4}}+{{\left( x+5 \right)}^{4}}=16$ is
(a) 0
(b) 2
(c) 4
(d) None of these

Answer 1

Hint: To find the roots of the given equation, factorize the given equation by splitting the middle terms. When the equation is factored up to a quadratic equation, calculate the discriminant of the equation to check the nature of the roots. Count all the real roots of the equation and ignore the imaginary ones.

Complete step-by-step answer:
We have to find the real roots of the equation ${{\left( x+3 \right)}^{4}}+{{\left( x+5 \right)}^{4}}=16$. To do so, we will firstly simplify the equation and then solve it to find the roots.
We know that ${{\left( a+b \right)}^{4}}={{a}^{4}}+{{b}^{4}}+4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}+4a{{b}^{3}}$.

$\Rightarrow$ ${{x}^{4}}+4{{x}^{3}}\left( 3 \right)+6{{x}^{2}}{{\left( 3 \right)}^{2}}+4x{{\left( 3 \right)}^{3}}+{{\left( 3 \right)}^{4}}+{{x}^{4}}+4{{x}^{3}}\left( 5 \right)+6{{x}^{2}}{{\left( 5 \right)}^{2}}+4x{{\left( 5 \right)}^{3}}+{{\left( 5 \right)}^{4}}=16$.
Simplifying the above expression, we have
$\Rightarrow$ $2{{x}^{4}}+32{{x}^{3}}+204{{x}^{2}}+608x+706=16$.
$\Rightarrow$ ${{x}^{4}}+16{{x}^{3}}+102{{x}^{2}}+304x+345=0$.
We will now factorize the above equation by splitting the terms.
Rearranging the terms of the above equation, we have
$\Rightarrow$ ${{x}^{4}}+3{{x}^{3}}+13{{x}^{3}}+39{{x}^{2}}+63{{x}^{2}}+189x+115x+345=0$.
Thus, we have
$\Rightarrow$ ${{x}^{3}}\left( x+3 \right)+13{{x}^{2}}\left( x+3 \right)+63x\left( x+3 \right)+115\left( x+3 \right)=0$.
Taking out the common terms, we have
$\Rightarrow$ $\left( x+3 \right)\left( {{x}^{3}}+13{{x}^{2}}+63x+115 \right)=0$.
Further splitting the terms of the equation, we have
$\Rightarrow$ $\left( x+3 \right)\left( {{x}^{3}}+5{{x}^{2}}+8{{x}^{2}}+40x+23x+115 \right)=0$.
Thus, we have
$\Rightarrow$ $\left( x+3 \right)\\{{{x}^{2}}\left( x+5 \right)+8x\left( x+5 \right)+23\left( x+5 \right)\\}=0$.
Taking out the common terms, we have $\left( x+3 \right)\left( x+5 \right)\left( {{x}^{2}}+8x+23 \right)=0$.
We will now split the terms of the equation ${{x}^{2}}+8x+23$.
We will firstly try to evaluate the discriminant of this equation.
We know that any equation of the form $a{{x}^{2}}+bx+c$ has the value of discriminant as ${{b}^{2}}-4ac$.
Substituting $a=1,b=8,c=23$ in the above equation, we have the value of discriminant as ${{\left( 8 \right)}^{2}}-4\left( 23 \right)=64-92=-28$.
We observe that the equation ${{x}^{2}}+8x+23$ has a negative value of discriminant. Thus, it has imaginary roots.
Hence, we observe that only real roots of the equation ${{\left( x+3 \right)}^{4}}+{{\left( x+5 \right)}^{4}}=16$ can be obtained by equating $\left( x+3 \right)\left( x+5 \right)$ to zero.
So, we have $\left( x+3 \right)\left( x+5 \right)=0$. Thus, we have $x=-3,-5$.
Hence, the equation ${{\left( x+3 \right)}^{4}}+{{\left( x+5 \right)}^{4}}=16$ has only two real roots, which is option (b).

Note: It’s not necessary to completely factorize the equation to find the number of real roots. We can check the nature of roots of a quadratic equation by calculating the value of discriminant. However, it’s not necessary to calculate the value of discriminant. We can factorize the equation completely and find its roots.