Question

How do you solve the equation $4{{x}^{2}}+4x=15$ using the quadratic formula?

Answer 1

Hint: Now we are given with a quadratic equation. We will first rearrange the terms to bring the equation in standard form $a{{x}^{2}}+bx+c=0$ . Now comparing with the standard form of the equation we get the values of a, b and c. Now we will use the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the roots of the equation.

Complete step-by-step solution:
Now first let us understand a quadratic equation.
Quadratic equation is an equation in which the highest degree of a variable is 2.
Hence the term ${{x}^{2}}$ must be present in a quadratic equation.
Now the general form of quadratic equation is $a{{x}^{2}}+bx+c=0$ . Where a, b and c are integers and x is called the variable.
Now the solution to any quadratic equation is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Here we can see that there are two solutions to the equation.
Now consider the given quadratic equation $4{{x}^{2}}+4x=15$
Now first let us rearrange the terms by bringing 15 on LHS hence, we get $4{{x}^{2}}+4x-15=0$
Comparing the given equation with the general form of quadratic equation $a{{x}^{2}}+bx+c=0$ we get, a = 4, b = 4 and c = - 15.
Now let us substitute the values in the formula for quadratic.
Hence we have
$\begin{align}
  & \Rightarrow x=\dfrac{-4\pm \sqrt{{{4}^{2}}-4\left( 4 \right)\left( -15 \right)}}{2\left( 4 \right)} \\
 & \Rightarrow x=\dfrac{-4\pm \sqrt{16+240}}{8} \\
 & \Rightarrow x=\dfrac{-4\pm \sqrt{256}}{8} \\
\end{align}$
$\Rightarrow x=\dfrac{-4\pm 16}{8}$
Hence we have $x=\dfrac{-4+16}{8}$ or $x=\dfrac{-4-16}{8}$
$\Rightarrow x=\dfrac{12}{8}$ or $x=\dfrac{-20}{8}$
Hence the roots of the given equation are $\dfrac{12}{8}$ and $\dfrac{-20}{8}$.

Note: Note that the nature of the roots of quadratic equation depends upon Discriminant $D=\sqrt{{{b}^{2}}-4ac}$ . If D > 0 then we get two real distinct roots. If D = 0 then we get real equal roots. If D < 0 then we get complex roots. Hence real roots are only possible for $D\ge 0$.