Question

How‌ ‌do‌ ‌you‌ ‌factor‌ ‌and‌ ‌solve:‌ ‌\[{{x}^{2}}+5x-2=0\]?

Answer 1

Hint: Consider the given quadratic polynomial equal to y. Now, use the completing the square method to factorize the given quadratic polynomial. Use the algebraic identities: - \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] to simplify the assumed expression ‘y’. Convert the given polynomial in the form \[\left( x-m \right)\left( x-n \right)\], where m and n are called zeroes of the polynomial. Substitute each term equal to 0 to find the values of x and get the roots.

Complete step-by-step solution:
Here, we have been provided with the quadratic equation: \[{{x}^{2}}+5x-2=0\] and we are asked to factorize it and then solve it.
Now, let us use the method of completing the square to factorize the given quadratic polynomial because here it will be difficult to use the middle term split method. So, using completing the square method, we get,
\[\begin{align}
  & \Rightarrow y={{x}^{2}}+5x-2 \\
 & \Rightarrow y={{x}^{2}}+2\times \dfrac{5}{2}\times x-2 \\
 & \Rightarrow y={{x}^{2}}+2\times \dfrac{5}{2}\times x+{{\left( \dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{5}{2} \right)}^{2}}-2 \\
\end{align}\]
On simplifying the above relation using the identity: - \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], we get,
\[\begin{align}
  & \Rightarrow y={{\left( x+\dfrac{5}{2} \right)}^{2}}-\left( \dfrac{25}{4}+2 \right) \\
 & \Rightarrow y={{\left( x+\dfrac{5}{2} \right)}^{2}}-\left( \dfrac{33}{4} \right) \\
\end{align}\]
We can write the above expression as: -
\[\Rightarrow y={{\left( x+\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{33}}{2} \right)}^{2}}\]
Using the algebraic identity: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], we get,
\[\Rightarrow y=\left( x+\dfrac{5}{2}+\dfrac{\sqrt{33}}{2} \right)\left( x+\dfrac{5}{2}-\dfrac{\sqrt{33}}{2} \right)\]
\[\Rightarrow y=\left[ x+\left( \dfrac{5+\sqrt{33}}{2} \right) \right]\left[ \left( x+\left( \dfrac{5-\sqrt{33}}{2} \right) \right) \right]\]
Hence, the above relation is the factored form of the given quadratic polynomial.
Now, to solve this polynomial we need to substitute the value of y equal to 0 and find the values of x. So, substituting y = 0, we get,
\[\Rightarrow \left( x+\left( \dfrac{5+\sqrt{33}}{2} \right) \right)\left( x+\left( \dfrac{5-\sqrt{33}}{2} \right) \right)=0\]
Substituting each term equal to 0, we get,
\[\Rightarrow x+\left( \dfrac{5+\sqrt{33}}{2} \right)=0\] or \[x+\left( \dfrac{5-\sqrt{33}}{2} \right)=0\]
\[\Rightarrow x=-\left( \dfrac{5+\sqrt{33}}{2} \right)\] or \[x=-\left( \dfrac{5-\sqrt{33}}{2} \right)\]
Hence, the above two values of x are the solutions of the given quadratic equation.

Note: One may note that here we have factored the polynomial first and the found the solutions. You can apply the reverse process also to get the required relations. What we can do is we will use the discriminant formula to find the two values of x first and then assume the solutions as x = m and x = n. In the final step of the solution we will consider the product \[\left( x-m \right)\left( x-n \right)\] to get the factored form.