Solveeit Logo
Login

Question

Question: How do you solve the equation \(4{{x}^{2}}+4x=15\) using the quadratic formula?...

How do you solve the equation 4x2+4x=154{{x}^{2}}+4x=15 using the quadratic formula?

Explanation

Solution

Now we are given with a quadratic equation. We will first rearrange the terms to bring the equation in standard form ax2+bx+c=0a{{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 −b±b2−4ac2a\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 x2{{x}^{2}} must be present in a quadratic equation.
Now the general form of quadratic equation is ax2+bx+c=0a{{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 −b±b2−4ac2a\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 4x2+4x=154{{x}^{2}}+4x=15
Now first let us rearrange the terms by bringing 15 on LHS hence, we get 4x2+4x−15=04{{x}^{2}}+4x-15=0
Comparing the given equation with the general form of quadratic equation ax2+bx+c=0a{{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
⇒x=−4±42−4(4)(−15)2(4) ⇒x=−4±16+2408 ⇒x=−4±2568 \begin{aligned} & \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{aligned}
⇒x=−4±168\Rightarrow x=\dfrac{-4\pm 16}{8}
Hence we have x=−4+168x=\dfrac{-4+16}{8} or x=−4−168x=\dfrac{-4-16}{8}
⇒x=128\Rightarrow x=\dfrac{12}{8} or x=−208x=\dfrac{-20}{8}
Hence the roots of the given equation are 128\dfrac{12}{8} and −208\dfrac{-20}{8}.

Note: Note that the nature of the roots of quadratic equation depends upon Discriminant D=b2−4acD=\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≥0D\ge 0.