Question

If \[{b^2} - 4ac \ge 0\] , then write the roots of a quadratic equation \[a{x^2} + bx + c = 0\] .

Answer 1

Hint: Here, we have to find the roots of the given quadratic equation where, \[{b^2} - 4ac \ge 0\]. We will use the method of completing the square and reach our desired answer by taking the square root on both sides.

Formula Used:
We will use the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] and \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\].

Complete step-by-step answer:
Given quadratic equation\[ = a{x^2} + bx + c = 0\]
Now, we know that the formula of Discriminant,\[D = {b^2} - 4ac\]
Also, according to the question,
\[{b^2} - 4ac \ge 0\]
\[ \Rightarrow D \ge 0\]
Now, before finding the roots of the given equation, we must know the formula and properties of Discriminant and we should keep in mind that,
We get two distinct roots i.e. positive and different when \[D > 0\].
We get two equal and positive roots when \[D = 0\].
And we get no real roots i.e. negative roots when \[D < 0\].
As it is given that,\[D \ge 0\], this means that roots are real and they can either be equal or distinct. This is given mainly to justify that the roots of the given equation exist.
Now, we will find the roots of the given equation,
\[a{x^2} + bx + c = 0\]
Taking \[c\] on the RHS, we get,
\[a{x^2} + bx = - c\]
Now, divide both sides by the coefficient of \[{x^2}\]. Therefore, we get
\[ \Rightarrow {x^2} + \dfrac{b}{a}x = \dfrac{{ - c}}{a}\]
Applying the method of completing the square, we get
\[ \Rightarrow {\left( x \right)^2} + \left( {\dfrac{b}{{2a}}} \right)2\left( x \right) + {\left( {\dfrac{b}{{2a}}} \right)^2} - {\left( {\dfrac{b}{{2a}}} \right)^2} = \dfrac{{ - c}}{a}\]
Simplifying further, we get,
\[{\left( {x + \dfrac{b}{{2a}}} \right)^2} - {\left( {\dfrac{b}{{2a}}} \right)^2} = \dfrac{{ - c}}{a}\]
Adding \[{\left( {\dfrac{b}{{2a}}} \right)^2}\]on both sides,
\[ \Rightarrow {\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{ - c}}{a} + {\left( {\dfrac{b}{{2a}}} \right)^2}\]
\[ \Rightarrow {\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{ - c}}{a} + \dfrac{{{b^2}}}{{4{a^2}}}\]
Taking LCM on RHS, we get,
\[ \Rightarrow {\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{{b^2} - 4ac}}{{4{a^2}}}\]
Now, taking square root on both sides,
\[ \Rightarrow \left( {x + \dfrac{b}{{2a}}} \right) = \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}} \]
We put both \[ \pm \] sign when we take a square root as we get both positive and negative square roots of the radicand (the number whose square root we are taking).
Subtracting \[\dfrac{b}{{2a}}\] on both sides,
\[ \Rightarrow x = \dfrac{{ - b}}{{2a}} \pm \dfrac{{\sqrt {{b^2} - 4ac} }}{{\sqrt {4{a^2}} }}\]
Now, taking the denominator common, we get,
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Or we can rewrite this as:
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
Hence, the required roots of the given quadratic equation \[a{x^2} + bx + c = 0\] are:
\[\dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}\] and \[\dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}\].

Note: The roots (of the general quadratic equation) which we have derived are used as Quadratic Formula. We can find roots of any quadratic equation, when it written in the form of $a{{x}^{2}}+bx+c=0$ by using this quadratic formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Completing the square is a method in which we try to make the identity of \[{\left( {a \pm b} \right)^2} = {a^2} + {b^2} \pm 2ab\] by adding and subtracting the second term. It is required to complete the identity and multiplying and dividing a given term (having $x$ as a variable) by 2 to make it is in the form \[2ab\] .