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Question: How do you solve the quadratic equation \(2{{x}^{2}}+7x=5\) ?...

How do you solve the quadratic equation 2x2+7x=52{{x}^{2}}+7x=5 ?

Explanation

Solution

In this problem, a quadratic equation is given which is in the form of ax2+bx+c=0a{{x}^{2}}+bx+c=0. For this particular equation, we have to find the roots or zeros. There are only two roots in a quadratic equation. Roots are the values which satisfy the equation and result in zero when roots are placed and solved. To find the roots, quadratic formula is used:
⇒x=−b±b2−4ac2a\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}

Complete step-by-step solution:
Now, let’s solve the problem.
The equation which possess the highest degree as 2 and which is in the form of ax2+bx+c=0a{{x}^{2}}+bx+c=0 is called quadratic equation where ‘a’ and ‘b’ are coefficients of x2{{x}^{2}} and x respectively, and c is the constant term. It has two roots which satisfy the equation and results in zero after placing the roots in the equation. And you should learn that there are basically two methods of finding roots and an equation. The two methods are middle term splitting and by using quadratic formula. So, we will use a quadratic formula for this equation because middle term splitting is only possible if we can split the middle term into two terms and we are getting something common while grouping them in pairs. Quadratic formula is:
⇒x=−b±b2−4ac2a\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Now, write the equation given in the question.
⇒2x2+7x=5 ⇒2x2+7x−5=0 \begin{aligned} & \Rightarrow 2{{x}^{2}}+7x=5 \\\ & \Rightarrow 2{{x}^{2}}+7x-5=0 \\\ \end{aligned}
Here, a = 2, b = 7 and c = -5. Place all the values in quadratic formula, we will get:
⇒x=−7±(7)2−4(2)(−5)2(2)\Rightarrow x=\dfrac{-7\pm \sqrt{{{\left( 7 \right)}^{2}}-4\left( 2 \right)\left( -5 \right)}}{2\left( 2 \right)}
On solving further:
⇒x=−7±49+404\Rightarrow x=\dfrac{-7\pm \sqrt{49+40}}{4}
Solve the under root:
⇒x=−7±894\Rightarrow x=\dfrac{-7\pm \sqrt{89}}{4}
Simplify further:
⇒x=−74±894\Rightarrow x=\dfrac{-7}{4}\pm \dfrac{\sqrt{89}}{4}
The roots for x are:
⇒x=−74−894,−74+894\Rightarrow x=\dfrac{-7}{4}-\dfrac{\sqrt{89}}{4},\dfrac{-7}{4}+\dfrac{\sqrt{89}}{4}
This is the final answer.

Note: Like other numbers, we haven’t taken the LCM of the under root value i.e. 89 because it is a prime number. We cannot take out anything from the root. So we leave the root as it is. At the first step itself, you have to write the equation in the form of ax2+bx+c=0a{{x}^{2}}+bx+c=0 before solving further.