Question

How do you solve the quadratic equation \[{{x}^{2}}+3x-28=0\] ?

Answer 1

Hint:Now to solve the given equation we will use the complete square method. Hence we will add the term ${{\left( \dfrac{b}{2a} \right)}^{2}}$ to the equation after making it monic. Now we will use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ to simplify the equation and then take square root after rearranging the terms. Now we will solve the obtained linear equations to find the value of x.

Complete step-by-step answer:
Now consider the given equation \[{{x}^{2}}+3x-28=0\]
Now we will solve this equation by using the complete square method.
Now first check if the coefficient of ${{x}^{2}}$ is one.
Now since the coefficient of the ${{x}^{2}}$ is already one then we can proceed further.
Now first we will add and subtract the equation by term ${{\left( \dfrac{b}{2a} \right)}^{2}}$ .
Hence we get,
$\Rightarrow {{x}^{2}}+3x+\dfrac{9}{4}-\dfrac{9}{4}-28$
Now we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ .
Hence we have $\Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}-28-\dfrac{9}{4}=0$
Now taking LCM we get,
$\begin{align}
  & \Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}+\dfrac{-28\times 4-9}{4}=0 \\
 & \Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}+\dfrac{-121}{4}=0 \\
 & \Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}=\dfrac{121}{4} \\
\end{align}$
Now taking the square root in both sides we get,
$\Rightarrow x+\dfrac{3}{2}=\pm \dfrac{11}{2}$
Now transposing $\dfrac{3}{2}$ on RHS we get, $x=\dfrac{11}{2}-\dfrac{3}{2}$ or $x=-\dfrac{11}{2}-\dfrac{3}{2}$ .
Hence we get, $x=4$ or $x=-7$
Hence the solutions of the equation are x = 4 and x = - 7.

Note: Now note that we can also directly solve this problem by using formula for roots of the quadratic equation. For any quadratic equation of the form $a{{x}^{2}}+bx+c=0$ we have the roots of the equation is given by formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . We can also use the method of splitting the middle term in which split the middle term such that the product of them is the multiplication of first term and last term. Hence we can take factors common from the first two terms and last two terms and solve the equation.