Question

Solve the given expression: ${{x}^{2}}+4x-8=0$.

Answer 1

Hint:Quadratic formula for finding the roots of the equation $A{{x}^{2}}+Bx+C=0$ is given as $x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$. Compare the given quadratic with $A{{x}^{2}}+Bx+C=0$ and get values of A, B and C and hence get the roots using the quadratic formula given above.

Complete step-by-step answer:
Given quadratic equation is
${{x}^{2}}+4x-8=0.................\left( i \right)$
As we know that we can solve any quadratic equation by factorizing it in two factors. So, the procedure factorizing any quadratic is that we need to split the middle term (coefficient of x) into two, such values such that multiplication of them will be equal to the multiplication of constant term and coefficient of ${{x}^{2}}$.
We know that we can solve any quadratic equation with the help of a quadratic formula given for calculating roots of any quadratic.
So, if we have any quadratic $A{{x}^{2}}+Bx+C=0$, then roots of this equation can be given by formula,
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}......................\left( ii \right)$
Now, we can find roots of the given quadratic equation using the equation (i). So, we need to compare the given equation by $A{{x}^{2}}+Bx+C=0$ to get values of A, B and C and hence, we can put the values of A, B and C in equation (ii) to get the roots of the given equation.
So, comparing equation (i) $1.{{x}^{2}}+4x-8$ with the equation $A{{x}^{2}}+Bx+C=0$, we get values of A, B and C as 1, 4 and -8 respectively. Hence, we get,
A = 1
B = 4
C = -8
Now, we can put the values of A, B and C in the equation (i) to get the roots of equation ${{x}^{2}}+4x-8=0$, we get,
$\begin{align}
  & x=\dfrac{-4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\times \left( 1 \right)\left( -8 \right)}}{2\times 1} \\
 & x=\dfrac{-4\pm \sqrt{16+32}}{2} \\
 & x=\dfrac{-4\pm \sqrt{48}}{2}........................\left( iii \right) \\
\end{align}$
Now, we can find the root of 48 by factorizing it. So, we get,
 $\begin{align}
  & 2\left| \!{\underline {\,
  48 \,}} \right. \\
 & 2\left| \!{\underline {\,
  24 \,}} \right. \\
 & 2\left| \!{\underline {\,
  12 \,}} \right. \\
 & 2\left| \!{\underline {\,
  6 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \ \ 1 \\
\end{align}$
So, we get,
$48=2\times2\times2\times2\times3$
Now, we can get roots by taking the same numbers in pairs and multiply it by once. Now 3 has no pair so, we will write it as $\sqrt{3}$, we get,
$\begin{align}
  & \sqrt{48}=2\times 2\times \sqrt{3} \\
 & \sqrt{48}=4\sqrt{3} \\
\end{align}$
Now, we can put $\sqrt{48}=4\sqrt{3}$ in the equation (iii) to get values of x, we get,
$x=\dfrac{-4\pm 4\sqrt{3}}{2}$
Now, we can take two cases, one by taking the ‘+’ in between ‘-4’ and $4\sqrt{3}$ and other by taking the ‘-‘ sign between them. Hence, we can get two values of x as,
Case 1:
$\begin{align}
  & x=\dfrac{-4+4\sqrt{3}}{2}=\dfrac{-4}{2}+\dfrac{4\sqrt{3}}{2} \\
 & x=-2+2\sqrt{3} \\
\end{align}$
Case 2:
\[\begin{align}
  & x=\dfrac{-4-4\sqrt{3}}{2}=\dfrac{-4}{2}-\dfrac{4\sqrt{3}}{2} \\
 & x=-2-2\sqrt{3} \\
\end{align}\]
Hence, two values of x can be given as $x=-2\pm 2\sqrt{3}$. So, roots of the given quadratic is $\left( -2+2\sqrt{3},-2-2\sqrt{3} \right)$
So, the answer is $\left( -2\pm 2\sqrt{3} \right)$.

Note: Quadratic formula can be proved as; we have $a{{x}^{2}}+bx+c=0$
\[\begin{align}
  & a\left[ {{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a} \right]=0 \\
 & \Rightarrow \left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}+\dfrac{c}{a}-\dfrac{{{b}^{2}}}{4{{a}^{2}}} \right]=0 \\
 & \Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}} \\
 & \Rightarrow x+\dfrac{b}{2a}=\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}} \\
 & \Rightarrow x=\dfrac{-b}{2a}\pm \dfrac{\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
\end{align}\]
Take care while putting values of a, b and c from the given quadratic to the quadratic formula. One may do mistake with any sign ( + or - ) as well. So, please take care of it as well.
Factorizing method to find roots of any quadratic will not work here. As we cannot split the middle term such that the product is equal to the product coefficient of ${{x}^{2}}$ and constant term. So, it is complex to guess the roots by this method.