Question

Solve the following equations: \[{{x}^{4}}-10{{x}^{2}}-20x-16=0\].

Answer 1

Hint: In order to find solution to this problem, we will first factorize our equation into simpler equation and then will use Zero Factor Principle to separate our equations and then will solve the equation as a quadratic equation and will simplify the equation by finding the roots and by identifying which formula to use.

Complete step-by-step solution:
We have our equation as:
\[\Rightarrow {{x}^{4}}-10{{x}^{2}}-20x-16=0\]
Now, first we will factorize an equation into a simpler equation.
On factoring our equation, we get:
$\Rightarrow \left( x+2 \right)\left( x-4 \right)\left( {{x}^{2}}+2x+2 \right)=0$
Now, we will use the Zero Factor Principle to separate our equations.
That is, If $ab=0$ then $a=0$or $b=0$
On applying Zero Factor Principle on our equation, we get:
$x+2=0\to \left( 1 \right)$ or $x-4=0\to \left( 2 \right)$ or ${{x}^{2}}+2x+2=0\to \left( 3 \right)$
Now, first we will solve the equation $\left( 1 \right)$.
That is, $x+2=0$
On simplifying, we get:
$\Rightarrow x=-2$
Then, we will solve the equation $\left( 2 \right)$.
That is, $x-4=0$
On simplifying, we get:
$\Rightarrow x=4$
Finally, we will solve the equation $\left( 3 \right)$.
That is, ${{x}^{2}}+2x+2=0$
Now since the above equation is quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ , we will use the following formula: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Therefore, from the above equation and on comparing, we get $a=1,b=2,c=2$
Now, substituting the values in formula we get: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 1\times 2}}{2\times 1}$
Now we will first simplify root part in our equation:
$\Rightarrow \sqrt{4-8}$
$\Rightarrow \sqrt{-4}$
Now by applying radical rule, i.e., $\sqrt{-a}=\sqrt{-1}\sqrt{a}$ we get:
$\Rightarrow \sqrt{-1}\sqrt{4}$
Now we can see we can apply imaginary rule: $\sqrt{-1}=i$
An imaginary number is a number that is expressed in terms of square root of a negative number (usually the square root of $-1$, and is represented by $i$ or $j$ )
Therefore, we get:
$\Rightarrow \sqrt{4}i$
On simplifying, we get:
$\Rightarrow 2i$
Now coming to our equation and substituting what we equate, we get:
$\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-2\pm 2i}{2}$
Now on factorization, we get:
$\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{2\left( -1\pm i \right)}{2}$
On dividing and equating, we get:
$\left( {{x}_{1}},{{x}_{2}} \right)=-1\pm i$
Now on equating the above equation we get the value of x, i.e.:
${{x}_{1}}=-1+i$ and ${{x}_{2}}=-1-i$
Therefore, with this, we get the solution of our following equation \[{{x}^{4}}-10{{x}^{2}}-20x-16=0\] as:
$x=-2$, $x=4$, $x=-1+i$, $x=-1-i$

Note: There are three forms of quadratic equation, Standard form:$y= a{{x}^{2}}+bx+c$ where $a,b,c$ are just numbers. Factored form:\[~y\text{ }=\text{ }\left( ax\text{ }+\text{ }c \right)\left( bx\text{ }+\text{ }d \right)\] again the $a,b,c,d$ are just numbers. Vertex form: \[y\text{ }=\text{ }a{{\left( x\text{ }+\text{ }b \right)}^{2}}\text{ }+\text{ }c\] again the $a,b,c$ are just numbers.
We have to identify by looking at the equation which form we have to use, then only we can solve the equation.
If we have a standard form of a quadratic equation, we can verify the answer by substituting the answer in the original equation.