Question

How do you solve ${{x}^{2}}+9x+9=0$ by using the quadratic formula ?

Answer 1

Hint: If the quadratic equation is in the form $a{{x}^{2}}+bx+c$ then the formula for roots of quadratic equation is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .By using this formula we can find the roots of ${{x}^{2}}+9x+9=0$ . Types of roots whether they will be real or distinct that depends on the discriminant ${{b}^{2}}-4ac$

Complete step by step answer:
We have to find the roots of the equation ${{x}^{2}}+9x+9=0$ . We can see that it is a quadratic equation.
The roots of quadratic equation $a{{x}^{2}}+bx+c$ are $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
If we compare ${{x}^{2}}+9x+9$ with $a{{x}^{2}}+bx+c$ then the value of a is 1 , value of b is 9 and value of c is 9
Roots of equation $a{{x}^{2}}+bx+c$ = $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
We can substitute the value of a, b and c
Roots of equation = $\dfrac{-9\pm \sqrt{{{9}^{2}}-4\times 1\times 9}}{2\times 1}$
Further solving we get
Roots of the equation = $\dfrac{-9\pm \sqrt{45}}{2}$
We can write $\sqrt{45}$ as $3\sqrt{5}$
So the roots are $\dfrac{-9\pm 3\sqrt{5}}{2}$
So one root is $-\dfrac{9}{2}-\dfrac{3}{2}\sqrt{5}$ and the other one is $-\dfrac{9}{2}+\dfrac{3}{2}\sqrt{5}$

Note:
Another method to find the roots of a quadratic equation is by factoring the equation. Sometimes the factoring method is easier but sometimes it is very difficult to factorize a quadratic equation in that case a quadratic formula is used to find the roots . in quadratic equation $a{{x}^{2}}+bx+c$ the sum of the roots is $-\dfrac{b}{a}$ and the product of roots is $\dfrac{c}{a}$ . If the discriminant ${{b}^{2}}-4ac$ is positive we get 2 distinct real roots, if the discriminant is negative we get 2 complex roots both roots are conjugate of each other. If the value of discriminant is 0 that means ${{b}^{2}}=4ac$ then both roots will be equal and real.