Question

How do you solve ${{x}^{2}}+2x-3=0$ using the quadratic formula?

Answer 1

Hint: From the given question we have to find the solutions for the ${{x}^{2}}+2x-3=0$ by using the quadratic formula. The quadratic formula helps us to solve any quadratic equation. First, we bring the equation to the form $a{{x}^{2}}+bx+c=0$, where a, b, and c are coefficients. Then, we plug these coefficients in the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. By using this formula, we will get the solutions for the above quadratic equation.

Complete step by step answer:
From the question given equation is
$\Rightarrow {{x}^{2}}+2x-3=0$
Now we have to solve this equation by using quadratic formula as we all know that,
The quadratic formula helps us to solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula
The formula is
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now the given quadratic equation is in the form of $a{{x}^{2}}+bx+c=0$ then by comparing the given equation with the general quadratic equation we will get the values of a, b and c.
By comparing we will get,
$\Rightarrow a=1$
$\Rightarrow b=2$
$\Rightarrow c=-3$
Now we have to substitute the above values in their respective positions in the formula.
By substituting the above values in their respective positions in the formula we will get,
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$\Rightarrow x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 1\times -3}}{2\times 1}$
$\Rightarrow x=\dfrac{-2\pm \sqrt{4+12}}{2}$
$\Rightarrow x=\dfrac{-2\pm \sqrt{16}}{2}$
Since the square root of 16 is 4, we get
$\Rightarrow x=\dfrac{-2\pm 4}{2}$
$\Rightarrow x=\dfrac{-2+4}{2}\quad or\quad x=\dfrac{-2-4}{2}$
$\Rightarrow x=\dfrac{2}{2}\quad or\quad x=\dfrac{-6}{2}$
$\Rightarrow x=1\quad or\quad x=-3$

Therefore, the solutions of the above equation ${{x}^{2}}+2x-3=0$ by using quadratic formula is $x=1\quad or\quad x=-3$.

Note: Students should know the basic formulas of quadratic equations. In the formula there is $\pm $ students should remember this if they only write plus they will only get one root. As in the above question they asked to do by using quadratic formulas. Without this also by factorization we can solve this equation. in the following way,
$\begin{align}
  & \Rightarrow {{x}^{2}}+2x-3=0 \\
 & \Rightarrow {{x}^{2}}+3x-x-3=0 \\
 & \Rightarrow x\left( x+3 \right)-\left( x+3 \right)=0 \\
 & \Rightarrow \left( x-1 \right)\left( x+3 \right)=0 \\
 & \Rightarrow x=1\quad or\quad x=-3 \\
\end{align}$