
What are the roots of the equation ${{x}^{2}}+4x-16=0$?
Answer
517.2k+ views
Hint: To obtain the roots of the given equation use Quadratic formula. Firstly we will write the general form of the quadratic formula and then compare it with the equation to get all the values. Then we will put the values in the formula and solve it to obtain the desired answer.
Complete step by step answer:
The equation is given as:
${{x}^{2}}+4x-16=0$……$\left( 1 \right)$
According to the quadratic formula for any equation of type below:
$a{{x}^{2}}+bx+c=0$…..$\left( 2 \right)$
Where $a,b,c$ are some constants
The value of variable $x$ is obtained by using the formula below:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$……$\left( 3 \right)$
Now on comparing equation (1) and (2) we get the values of $a,b,c$ as:
$\begin{align}
& a=1 \\
& b=4 \\
& c=-16 \\
\end{align}$
Substituting the above value in equation (3) and simplifying we get,
$\begin{align}
& x=\dfrac{-\left( 4 \right)\pm \sqrt{{{\left( 4 \right)}^{2}}-4\times 1\times \left( -16 \right)}}{2\times 1} \\
& \Rightarrow x=\dfrac{-4\pm \sqrt{16+64}}{2} \\
& \Rightarrow x=\dfrac{-4\pm \sqrt{80}}{2} \\
\end{align}$
$\therefore x=\dfrac{-4\pm 4\sqrt{5}}{2}$
So we got the two values as:
$\begin{align}
& x=\dfrac{-4+4\sqrt{5}}{2} \\
& \Rightarrow x=\dfrac{2\left( -2+2\sqrt{5} \right)}{2} \\
& \therefore x=-2+2\sqrt{5} \\
\end{align}$
And
$\begin{align}
& x=\dfrac{-4-4\sqrt{5}}{2} \\
& \Rightarrow x=\dfrac{2\left( -2-2\sqrt{5} \right)}{2} \\
& \therefore x=-2-2\sqrt{5} \\
\end{align}$
So the values obtained are $x=-2+2\sqrt{5}$ and $x=-2-2\sqrt{5}$
Hence roots of equation ${{x}^{2}}+4x-16=0$ are $x=-2+2\sqrt{5}$ and $x=-2-2\sqrt{5}$.
Note: Quadratic formula is used to find the solution of a quadratic equation. It is easy and less complicated than other methods such as completing the square, factoring by inspection or using Geometrical Interpretation. One thing is to be kept in mind though while using this formula that the value of the constant is taken correctly. This formula is straightforward and is used when the equation is quadratic in nature. Roots of an equation are those values which when substituted in the equation satisfy the equation that is it makes the left side equal to the right side. The roots can be real as well as complex in nature. A quadratic equation always has two roots which can be real or complex in nature.
Complete step by step answer:
The equation is given as:
${{x}^{2}}+4x-16=0$……$\left( 1 \right)$
According to the quadratic formula for any equation of type below:
$a{{x}^{2}}+bx+c=0$…..$\left( 2 \right)$
Where $a,b,c$ are some constants
The value of variable $x$ is obtained by using the formula below:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$……$\left( 3 \right)$
Now on comparing equation (1) and (2) we get the values of $a,b,c$ as:
$\begin{align}
& a=1 \\
& b=4 \\
& c=-16 \\
\end{align}$
Substituting the above value in equation (3) and simplifying we get,
$\begin{align}
& x=\dfrac{-\left( 4 \right)\pm \sqrt{{{\left( 4 \right)}^{2}}-4\times 1\times \left( -16 \right)}}{2\times 1} \\
& \Rightarrow x=\dfrac{-4\pm \sqrt{16+64}}{2} \\
& \Rightarrow x=\dfrac{-4\pm \sqrt{80}}{2} \\
\end{align}$
$\therefore x=\dfrac{-4\pm 4\sqrt{5}}{2}$
So we got the two values as:
$\begin{align}
& x=\dfrac{-4+4\sqrt{5}}{2} \\
& \Rightarrow x=\dfrac{2\left( -2+2\sqrt{5} \right)}{2} \\
& \therefore x=-2+2\sqrt{5} \\
\end{align}$
And
$\begin{align}
& x=\dfrac{-4-4\sqrt{5}}{2} \\
& \Rightarrow x=\dfrac{2\left( -2-2\sqrt{5} \right)}{2} \\
& \therefore x=-2-2\sqrt{5} \\
\end{align}$
So the values obtained are $x=-2+2\sqrt{5}$ and $x=-2-2\sqrt{5}$
Hence roots of equation ${{x}^{2}}+4x-16=0$ are $x=-2+2\sqrt{5}$ and $x=-2-2\sqrt{5}$.
Note: Quadratic formula is used to find the solution of a quadratic equation. It is easy and less complicated than other methods such as completing the square, factoring by inspection or using Geometrical Interpretation. One thing is to be kept in mind though while using this formula that the value of the constant is taken correctly. This formula is straightforward and is used when the equation is quadratic in nature. Roots of an equation are those values which when substituted in the equation satisfy the equation that is it makes the left side equal to the right side. The roots can be real as well as complex in nature. A quadratic equation always has two roots which can be real or complex in nature.
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