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Let \[a > 0,\,b > 0\] and \[c > 0\], then, what are both the roots of the equation \[a{x^2} + bx + c = 0\]?

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Last updated date: 12th Jul 2024
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Answer
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Hint: Consider the various cases of the discriminant of the quadratic equation and determine the nature of the roots for each case. 

 

Complete step-by-step answer:

We know that for a quadratic equation given by $ax^2+bx+c=0$, the roots are given by the quadratic formula as, $x={\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}}$

In the above formula $b^2-4ac$ is known as the discriminant (D) and it decides the nature of the roots.

When D > 0, the roots are real and different.

When D = 0, the roots are real and equal.

When D < 0, the roots are imaginary.

The quadratic formula can be rewritten as $x={\dfrac{-b \pm \sqrt{D}}{2a}}$

In this problem, we are given that $a > 0$, $b > 0$, $c > 0$. Based on this, we can write

$\Rightarrow b^2 > b^2 - 4ac$

Applying square root on both sides, we get,

$\Rightarrow |b| > \sqrt{b^2 - 4ac}$

But b > 0, so |b| = b,

$\Rightarrow b > \sqrt{b^2 - 4ac}$

$\therefore$ The numerator of the quadratic formula $(-b \pm \sqrt{b^2-4ac})$ will always be negative in this problem.

Let us discuss all the cases:

Case I: D > 0

The roots are $x = {\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}}$

Both are real, different and negative.

 

Case II: D = 0

The roots are $x = {\dfrac{-b}{2a}}$

Both are real, equal and negative.

 

Case III: D < 0

The roots are $x = {\dfrac{-b \pm i \sqrt{4ac - b^2}}{2a}}$

Both are complex with negative real parts.

 

Note: Such problems can be solved by using the concept of discriminant in the quadratic formula and the conditions given in the problem statement. Mistakes can be avoided while considering the inequalities.