Question

Find all the possible values of x such that the following equation holds:
${{x}^{2}}-6x+7=0$

Answer 1

Hint: This is a quadratic equation. So the Sridharacharya formula can be applied to solve this equation for x. For a quadratic equation of the form:
$a{{x}^{2}}+bx+c=0\text{ },\text{ }a\ne 0\text{ }\cdot \cdot \cdot \text{(i)}$
The roots of the equation (i) can be determined by using Sridharacharya formula as follow:
$\begin{align}
 & {{x}_{1}}=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & {{x}_{2}}=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \\
\end{align}$


Complete step-by-step answer:
A quadratic equation can be solved using the Sridharacharya formula. According to this formula if a quadratic equation is of the form:
$a{{x}^{2}}+bx+c=0\text{ },\text{ }a\ne 0\text{ }\cdot \cdot \cdot \text{(i)}$
Then the roots of this equation can be determined by
$\begin{align}
 & {{x}_{1}}=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & {{x}_{2}}=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \\
\end{align}$
First let us prove the Shridharacharya formula by solving equation (i).
$a{{x}^{2}}+bx+c=0$
Multiplying $a$ throughout the equation we get
$\begin{align}
 & {{a}^{2}}{{x}^{2}}+abx+ac=0 \\
 & \Rightarrow {{a}^{2}}{{x}^{2}}+2\dfrac{ab}{2}x+ac=0 \\
 & \Rightarrow {{a}^{2}}{{x}^{2}}+2\dfrac{ab}{2}x+\dfrac{{{b}^{2}}}{4}-\dfrac{{{b}^{2}}}{4}+ac=0 \\
 & \Rightarrow {{\left( ax \right)}^{2}}+2\left( ax \right)\left( \dfrac{b}{2} \right)+{{\left( \dfrac{b}{2} \right)}^{2}}=\dfrac{{{b}^{2}}}{4}-ac \\
\end{align}$

We see that LHS forms a square expression. So converting it into square form we get
$\begin{align}
 & {{\left( ax+\dfrac{b}{2} \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{4} \\
 & \Rightarrow \left( ax+\dfrac{b}{2} \right)=\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{4}}=\pm \dfrac{\sqrt{{{b}^{2}}-4ac}}{2}\text{ }\cdots \text{(ii)} \\
\end{align}$

We know that the square root of negative numbers is not defined. So for the equation (i) to have real roots, the term inside the square root in the above expression should be non-negative. This term is called discriminant of the quadratic equation.
${{b}^{2}}-4ac\ge 0\text{ }\cdots \text{(iii)}$
Equation (iii) is a condition for equation (i) to have real roots. Also when (iii) holds then we will have two square roots, one positive and another negative.

Now solving further from equation (ii),
$\begin{align}
 & \left( ax+\dfrac{b}{2} \right)=\pm \dfrac{\sqrt{{{b}^{2}}-4ac}}{2} \\
 & \Rightarrow ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2} \\
 & \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
\end{align}$
Considering both the sign + and – we have two roots as:
$\begin{align}
 & {{x}_{1}}=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & {{x}_{2}}=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \\
\end{align}$

In this case, there is one special case when both the roots are the same and the equation has only one distinct root. This special case occurs when
${{b}^{2}}-4ac=0$
Putting the above equation in the expression of x1 and x2 we get,
${{x}_{1}}={{x}_{2}}=\dfrac{-b}{2a}$

If ${{b}^{2}}-4ac<0$ then square root doesn’t exist over real numbers and hence no real root x of equation (i) would be possible. In this case, complex roots occur.
$\begin{align}
 & {{b}^{2}}-4ac<0 \\
 & \Rightarrow 4ac-{{b}^{2}}>0 \\
\end{align}$
Solving the equation for this case from equation (ii),
$\begin{align}
 & \left( ax+\dfrac{b}{2} \right)=\pm \dfrac{\sqrt{{{b}^{2}}-4ac}}{2} \\
 & \Rightarrow \left( ax+\dfrac{b}{2} \right)=\pm \dfrac{\sqrt{-\left( 4ac-{{b}^{2}} \right)}}{2} \\
 & \Rightarrow \left( ax+\dfrac{b}{2} \right)=\pm \dfrac{\sqrt{-1}\sqrt{\left( 4ac-{{b}^{2}} \right)}}{2} \\
\end{align}$
Now taking an imaginary number $i=\sqrt{-1}$ . Since $4ac-{{b}^{2}}>0$so $\sqrt{4ac-{{b}^{2}}}$exist.
$\begin{align}
 & \left( ax+\dfrac{b}{2} \right)=\pm \dfrac{i\sqrt{4ac-{{b}^{2}}}}{2} \\
 & \Rightarrow ax=\dfrac{-b\pm i\sqrt{4ac-{{b}^{2}}}}{2} \\
 & \Rightarrow x=\dfrac{-b\pm i\sqrt{4ac-{{b}^{2}}}}{2a} \\
\end{align}$
Again considering both the + and – sign we get two complex roots.
$\begin{align}
 & {{x}_{1}}=\dfrac{-b-i\sqrt{4ac-{{b}^{2}}}}{2a} \\
 & {{x}_{2}}=\dfrac{-b+i\sqrt{4ac-{{b}^{2}}}}{2a} \\
\end{align}$
We are given the equation as:
${{x}^{2}}-6x+7=0$
First let us check whether this equation has real roots or not.
Comparing this equation with equation (i) we get,
$a=1\text{ },\text{ }b=-6\text{ },\text{ }c=7$
Now checking the discriminant,
$\begin{align}
 & {{b}^{2}}-4ac={{\left( -6 \right)}^{2}}-4\left( 1 \right)\left( 7 \right) \\
 & \Rightarrow {{b}^{2}}-4ac=36-28 \\
 & \Rightarrow {{b}^{2}}-4ac=8\ge 0 \\
\end{align}$
We see that the discriminant is positive. So our equation has real roots. Now solving it using Shridharacharya formula we get
 $\begin{align}
 & {{x}_{1}}=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow {{x}_{1}}=\dfrac{-\left( -6 \right)-\sqrt{8}}{2\cdot 1} \\
 & \Rightarrow {{x}_{1}}=\dfrac{6-2\sqrt{2}}{2} \\
 & \Rightarrow {{x}_{1}}=3-\sqrt{2} \\
\end{align}$
And
$\begin{align}
 & {{x}_{2}}=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow {{x}_{2}}=\dfrac{-\left( -6 \right)+\sqrt{8}}{2\cdot 1} \\
 & \Rightarrow {{x}_{2}}=\dfrac{6+2\sqrt{2}}{2} \\
 & \Rightarrow {{x}_{2}}=3+\sqrt{2} \\
\end{align}$
So we get the roots of the equation
${{x}^{2}}-6x+7=0$
As ${{x}_{1}}=3-\sqrt{2}\text{ , }{{\text{x}}_{2}}=3+\sqrt{2}$

Note: You can save your time by first checking whether the discriminant is negative or non-negative. If it is non-negative then only proceed to solve otherwise you can directly say that this equation has no real roots.