Question

How to use the discriminant to find out how many real number roots an equation has for ${a^2} + 12a + 36 = 0$?

Answer 1

Hint: First compare the given quadratic equation to standard quadratic equation and find the value of numbers $A$, $b$ and $c$ in given equation. Then, substitute the values of $A$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, use the value of discriminant to determine the number of real roots of a given equation.

Formula used:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$……(i)
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.

Complete step by step solution:
First, we have to compare the given quadratic equation to the standard quadratic equation and find the value of numbers $A$, $b$ and $c$.
Comparing ${a^2} + 12a + 36 = 0$ with $A{x^2} + bx + c = 0$, we get
$A = 1$, $b = 12$ and $c = 36$
Now, we have to substitute the values of $A$, $b$ and $c$ in $D = {b^2} - 4Ac$ and find the discriminant of the given equation.
$D = {\left( {12} \right)^2} - 4\left( 1 \right)\left( {36} \right)$
After simplifying the result, we get
$ \Rightarrow D = 144 - 144$
$ \Rightarrow D = 0$
Since, the discriminant of the equation is zero, then the equation has real and equal roots.

Final solution: Hence, the given equation has two real and equal roots.

Note: Let $a{x^2} + bx + c = 0$, $a \ne 0$ be a quadratic equation. Then, the roots of this equation are given by
$\alpha = \dfrac{{ - b + \sqrt D }}{{2a}}$ and $\beta = \dfrac{{ - b - \sqrt D }}{{2a}}$.
If $D = {b^2} - 4ac > 0$, then $\alpha $ and $\beta $ are real.
Also, $\alpha - \beta = \left( {\dfrac{{ - b + \sqrt D }}{{2a}}} \right) - \left( {\dfrac{{ - b - \sqrt D }}{{2a}}} \right)$
$ \Rightarrow \alpha - \beta = \dfrac{{ - b + \sqrt D + b + \sqrt D }}{{2a}}$
$ \Rightarrow \alpha - \beta = \dfrac{{2\sqrt D }}{{2a}}$
$ \Rightarrow \alpha - \beta = \dfrac{{\sqrt D }}{a}$
$ \Rightarrow \alpha - \beta \ne 0$
$ \Rightarrow \alpha \ne \beta $
Thus, if $D = {b^2} - 4ac > 0$, i.e., the discriminant of the equation is positive, then the equation has real and distinct roots $\alpha $ and $\beta $ given by
$\alpha = \dfrac{{ - b + \sqrt D }}{{2a}}$ and $\beta = \dfrac{{ - b - \sqrt D }}{{2a}}$
If $D = {b^2} - 4ac = 0$, i.e., then $\alpha $ and $\beta $ are real.
Putting $D = 0$ in the expression for $\alpha $ and $\beta $.
$\alpha = - \dfrac{b}{{2a}} = \beta $
Thus, if $D = {b^2} - 4ac = 0$, i.e., the discriminant of the equation is zero, then the equation has real and equal roots equal to $ - \dfrac{b}{{2a}}$.
If $D = {b^2} - 4ac < 0$, i.e., the discriminant of the equation is negative, then the equation has no real roots.