Question

if $ {b^2} - 4ac \geqslant 0, $ then write the roots of a quadratic equation $ a{x^2} + bx + c = 0 $

Answer 1

Hint:
A quadratic equation in the variable x is an equation of the form $ a{x^2} + bx + c = 0 $ , for example, $ 2{x^2} + x - 300 = 0 $ is a quadratic equation.
In fact, any equation of the form $ p(x) = 0, $ where $ p(x) $ is a polynomial of degree 2 is a quadratic equation. But when we write the terms of $ p(x) $ in descending order of their degrees, then we get the standard form of the equation.

Complete step by step solution:
Step1
Given equation $ a{x^2} + bx + c = 0 $ is in one variable x in which $ a \ne 0\& 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 the 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. when we put the value of x in the given equation, it becomes zero.
In the given equation $ a{x^2} + bx + c = 0 $ only one condition is given $ {b^2} - 4ac \geqslant 0, $
Step 2
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 terms a cancel by the coefficient of $ {x^2} $ . So, using the completing square method. take half of the coefficient of x then add and subtract the squares in the equations.
 $ {x^2}\dfrac{b}{a}x + {\left( {\dfrac{b}{{2a}}} \right)^2} - {\left( {\dfrac{b}{{2a}}} \right)^2} + \dfrac{c}{a} = 0 $ .now, we use the identity $ {(a + b)^2} = {a^2} + {b^2} + 2ab $
We get, $ {\left( {x + \dfrac{b}{{2a}}} \right)^2} - \dfrac{{{b^2}}}{{4{a^2}}} + \dfrac{c}{a} = 0 $
This is the same as $ {\left( {x + \dfrac{b}{{2a}}} \right)^2} - \dfrac{{{b^2} - 4ac}}{{4{a^2}}} = 0 $
Take the second term to the right-hand side of the equation 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
If $ {b^2} - 4ac \geqslant 0, $ then by taking the square roots, we get
\[\left( {x + \dfrac{b}{{2a}}} \right) = \dfrac{{\sqrt {{b^2} - 4ac} }}{{2a}}\]
Therefore, \[x = \dfrac{{ - b}}{{2a}} + \dfrac{{\sqrt {{b^2} - 4ac} }}{{2a}}\] by lcm, we get \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

So, the roots of the equation are \[x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}\]and \[x = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}\]

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