Question

Show that the equation $9{{x}^{2}}+7x-2=0$ has real roots.

Answer 1

Hint: We know that the discriminant, for a quadratic equation $a{{x}^{2}}+bx+c=0$, is defined as $D={{b}^{2}}-4ac$, and the roots are real when the discriminant is non negative. We must first find the discriminant to show that it is non negative. Then, we also need to find the roots to verify the same.

Complete step by step answer:
We know that for a quadratic equation of the form $a{{x}^{2}}+bx+c=0$, discriminant is defined as follows, $D={{b}^{2}}-4ac$.
We also know that if the value of discriminant D is greater than or equal to zero, that is, $D\ge 0$ then the quadratic equation has real roots. And, if the discriminant D is negative, that is $D<0$, then the quadratic equation has complex roots.
We are also aware that the roots of this quadratic equation $\alpha \text{ and }\beta $ is given by,
$\alpha =\dfrac{-b+\sqrt{D}}{2a},\text{ }\beta =\dfrac{-b-\sqrt{D}}{2a}$.
Here, we have the quadratic equation $9{{x}^{2}}+7x-2=0$.
On comparing this quadratic equation with the general form $a{{x}^{2}}+bx+c=0$, we can say that
a = 9, b = 7 and c = -2.
Let us find the value of discriminant for this quadratic equation. We get
$D={{7}^{2}}-4\left( 9 \right)\left( -2 \right)$
On simplification, we get
$D=49+72$
Hence, we have
$D=121$
Now, we can easily see that $D>0$. Hence, we can say that the quadratic equation will have real roots.
We now have to verify this by finding the roots for this quadratic equation.
Let $\alpha \text{ and }\beta $ be the roots of this quadratic equation. Then, by using the formula given above, we can write
 $\alpha =\dfrac{-7+\sqrt{121}}{2\times 9},\text{ }\beta =\dfrac{-7-\sqrt{121}}{2\times 9}$.
Thus, we get
$\alpha =\dfrac{-7+11}{18},\text{ }\beta =\dfrac{-7-11}{18}$
Upon further simplification, we can write
$\alpha =\dfrac{4}{18},\text{ }\beta =\dfrac{-18}{18}$
On cancelling the common terms from numerator and denominator, we get
$\alpha =\dfrac{2}{9},\text{ }\beta =-1$.
Hence, we can see that both the roots of this quadratic equation are real.

Note: We must remember that the condition $D\ge 0$ for real roots, holds true with the combination of another condition that is $a\ne 0$. This extra condition is important to make sure that the given equation is a quadratic equation.