Question

How do you solve $2{x^2} = 14x + 20$ ?

Answer 1

Hint: In this question, they have given an equation and asked us to factor and solve it. First we need to alter the given equation by transferring all the terms to one side. Then we will solve the given equation using the quadratic equation formula. On doing some simplification we get the required answer.

Formula used: Quadratic equation formula:
 $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
The given equation is $2{x^2} = 14x + 20$ and we need to solve this equation.
First we should alter the given equation by transferring all the numbers and variables to the one side.
$2{x^2} = 14x + 20$
$2{x^2} - 14x - 20 = 0$
We can divide the equation by $2$ ,
$2{x^2} - 14x - 20 = 0$
${x^2} - 7x - 10 = 0$
Quadratic equation formula:
 $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Rewriting the given equation we get, \[{x^2} - 7x - 10 = 0\]
Clearly in\[{x^2} - 7x - 10 = 0\],
${\text{a = 1}}$,
${\text{b = - 7}}$,
${\text{c = - 10}}$.
Now applying \[{x^2} - 7x - 10 = 0\]in$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$,
  $ \Rightarrow x = \dfrac{{ - ( - 7) \pm \sqrt {{{( - 7)}^2} - 4(1)( - 10)} }}{{2(1)}}$
Simplifying the numerator we get,
$ \Rightarrow x = \dfrac{{7 \pm \sqrt {49 + 40} }}{2}$
$ \Rightarrow x = \dfrac{{7 \pm \sqrt {89} }}{2}$
Now we have to expand the expression into two, as there is a $ \pm $ in the expression. One becomes plus and the other becomes minus.
 $ \Rightarrow x = \dfrac{{7 + \sqrt {89} }}{2},x = \dfrac{{7 - \sqrt {89} }}{2}$

Therefore the roots are $x = \dfrac{{7 + \sqrt {89} }}{2},x = \dfrac{{7 - \sqrt {89} }}{2}$.

Note: Whenever the question asks you to find the roots with a quadratic equation, first you need to check the roots whether the roots exist or not and if it exists is it clear roots or equal roots.
 To check the roots, we have to find the value of $\sqrt {{b^2} - 4ac} $ first,
If $\sqrt {{b^2} - 4ac} $ is greater than $0$ , then the roots are different and clear.
If $\sqrt {{b^2} - 4ac} $ is equal to $0$, then the roots are equal.
If $\sqrt {{b^2} - 4ac} $ is less than $0$, then the roots are not real or real roots do not exist.