Question

How do you solve $2{{x}^{2}}+5x+5=0$ using the quadratic formula?

Answer 1

Hint: Write down the formula for finding the roots for a quadratic equation and compare the variables in the formula with that of the given quadratic equation. Now substitute the values in the formula to get two roots for the equation. The roots will be the solution for the given quadratic expression. Check if the roots formed then are real or imaginary.

Complete step-by-step solution:
The given expression is, $2{{x}^{2}}+5x+5=0$
The roots of a quadratic equation( $a{{x}^{2}}+bx+c=0$ ) can be found by the formula,
$x=\left[ \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \right]$
Here,$a=2;b=5;c=5$
Substitute the values in the formula.
$\Rightarrow x=\left[ \dfrac{-5\pm \sqrt{{{5}^{2}}-4\times 2\times 5}}{2\times 2} \right]$
Evaluate the expression under the square root first.
$\Rightarrow x=\left[ \dfrac{-5\pm \sqrt{25-40}}{2\times 2} \right]$
Since our ${{b}^{{}}}-4ac$ is $<0$; The roots hence found will be imaginary that is a complex number.
$\Rightarrow x=\left[ \dfrac{-5\pm \sqrt{\left( -15 \right)}}{2\times 2} \right]$
There is a notation for $\sqrt{(-1)}$ which is denoted by $i$ .
Hence after substituting the notation, our answer will now be,
$\Rightarrow x=\left[ \dfrac{-5\pm \sqrt{15}i}{4} \right]$
Now split open the $\pm$ operation and write two separate solutions.
$\Rightarrow x=\left[ \dfrac{-5+\sqrt{15}i}{4} \right];\left[ \dfrac{-5-\sqrt{15}i}{4} \right]$
Hence our roots for the quadratic expression are imaginary and complex.
$\therefore$ The solution for the quadratic equation $2{{x}^{2}}+5x+5=0$ is $x=\left[ \dfrac{-5+\sqrt{15}i}{4} \right];\left[ \dfrac{-5-\sqrt{15}i}{4} \right]$

Note: A polynomial is a mathematical expression which contains one or more variables in sum or subtraction format with different powers. Whenever there is a polynomial that is to be solved, the solution contains the roots of the expression. The number of roots is decided by the degree of the polynomial.
Never forget to take two conditions whenever there is a $\pm$ symbol. The expression is written twice once with $+\;$ and another time with $-\;$ . You can always cross-check your answer by placing the value of $x$ back in the equation. If you get LHS = RHS then your answer is correct.