Question

If the value of ${b^2} - 4ac$ is greater than zero, the quadratic equation $a{x^2} + bx + c = 0$ will have
A. Two Equal Real Roots
B. Two Distinct Real Roots
C. No Real Roots
D. No Roots or Solutions

Answer 1

An example of quadratic equation is $2{x^2} + x - 300 = 0$. In fact, any equation of the form $p\left( x \right) = 0$, where $p\left( x \right)$ is a polynomial of degree $2$ is a quadratic equation. But when we write the terms of $p\left( x \right)$ in descending order of their degrees, then we get the standard form of the equation.

Complete step-by-step answer:
Step 1: Given quadratic equation:
$a{x^2} + bx + c = 0$
Given a quadratic equation has one-variable $x$ in which $a \ne 0$, and $b,c \geqslant 0$.
$a$ is the coefficient of ${x^2}$, $b$ is the coefficient of $x$ and $c$ is the constant term.
Here, $a,b$ and $c$ are real numbers.
We have to find the roots of the given equation $a{x^2} + bx + c = 0$
i.e., we have to find the two values of $x$ such that when we put them in the given equation, we get the value of the equation equal to zero.
In this question, we have been given one condition: ${b^2} - 4ac \geqslant 0$.
Step 2: So, consider the quadratic equation $a{x^2} + bx + c = 0$ where $a \ne 0$.
Dividing throughout (both sides) by $a$, we get:
${x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0$
Here, in first term, $a$ is cancelled by the coefficient of ${x^2}$
So, using the completing square method, take the half of the coefficient of x, then add and subtract the squares in the equation.
${x^2} + \dfrac{b}{a}x + {\left( {\dfrac{b}{{2a}}} \right)^2} - {\left( {\dfrac{b}{{2a}}} \right)^2} + \dfrac{c}{a} = 0$
We will use the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
We get, ${\left( {x + \dfrac{b}{{2a}}} \right)^2} - \dfrac{{{b^2}}}{{4{a^2}}} + \dfrac{c}{a} = 0$
This is same as:
${\left( {x + \dfrac{b}{{2a}}} \right)^2} - \dfrac{{{b^2} - 4ac}}{{4{a^2}}} = 0$
Take the second term of the equation to the right-hand side, we get:
${\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{{b^2} - 4ac}}{{4{a^2}}}$
So, the roots of the given equation are the same as those of ${\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{{b^2} - 4ac}}{{4{a^2}}}$
Step 3: ${b^2} - 4ac \geqslant 0$, then by taking the square roots, we get
$x + \dfrac{b}{{2a}} = \pm \dfrac{{\sqrt {{b^2} - 4ac} }}{{2a}}$, here, $\sqrt {4{a^2}} = 2a$
Therefore, $x = \dfrac{{ - b}}{{2a}} + \dfrac{{\sqrt {{b^2} - 4ac} }}{{2a}}$
By the LCM, we get
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, roots of $a{x^2} + bx + c = 0$ are:
$x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$

Hence, it has two distinct real roots, so option (B) is correct.

Note: If ${b^2} - 4ac \geqslant 0$, then the roots of the quadratic equation $a{x^2} + bx + c = 0$ are given by $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. This formula for finding the roots of a quadratic equation is known as the quadratic formula. If ${b^2} - 4ac \geqslant 0$, then there is no real number whose square is ${b^2} - 4ac$. Therefore, there are no real roots for the given quadratic equation in this case.