Question

How do you solve ${{x}^{2}}+2x-28=0$?

Answer 1

Hint: We keep the variables and the constants separate on both sides of the equality. Then we add 1 to both sides of the equation ${{x}^{2}}+2x=28$. We then form a square for the left side of the new equation. Then we take the square root on both sides of the equation. From that we subtract 1 to the both sides to find the value of $x$ for ${{x}^{2}}+2x=28$.

Complete step by step solution:
We need to find the solution of the given equation ${{x}^{2}}+2x-28=0$.
We separate the variables and the constants on both sides of the equality.
We first add 1 to both sides of ${{x}^{2}}+2x=28$. We use the identity ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$.
$\begin{align}
  & {{x}^{2}}+2x+1=28+1 \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}=29 \\
\end{align}$
We interchanged the numbers for $a=x,b=1$.
Now we have a quadratic equation ${{\left( x+1 \right)}^{2}}=29$.
We need to find the solution of the given equation ${{\left( x+1 \right)}^{2}}=29$.
We take square root on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
$\begin{align}
  & \sqrt{{{\left( x+1 \right)}^{2}}}=\pm \sqrt{29} \\
 & \Rightarrow \left( x+1 \right)=\pm \sqrt{29} \\
\end{align}$
Now we subtract 1 to the both sides of the equation $\left( x+1 \right)=\pm \sqrt{29}$ to get value for variable $x$.
$\begin{align}
  & \left( x+1 \right)-1=\pm \sqrt{29}-1 \\
 & \Rightarrow x=-1\pm \sqrt{29} \\
\end{align}$
The given quadratic equation has two solutions and they are $x=-1\pm \sqrt{29}$.

Note: The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have two roots. Cubic polynomials have three. It can be both real and imaginary roots.
We can also apply the quadratic equation formula to solve the equation ${{x}^{2}}+2x-28=0$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have ${{x}^{2}}+2x-28=0$. The values of a, b, c is $1,2,-28$ respectively.
$x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 1\times \left( -28 \right)}}{2\times 1}=\dfrac{-2\pm \sqrt{116}}{2}=\dfrac{-2\pm 2\sqrt{29}}{2}=-1\pm \sqrt{29}$.
The given quadratic equation has two solutions and they are $x=-1\pm \sqrt{29}$.