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# 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

Last updated date: 14th Jul 2024
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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.

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.