Question

How do you solve ${{\text{x}}^2} = 5{\text{x + 10}}$ using the quadratic equation formula?

Answer 1

Hint: In this question, we are asked to solve the given equation using the quadratic equation formula. The quadratic equation formula is $\dfrac{{{{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$ .
First, we have to check the existence of roots using the formula stated below and find the types of roots. After that, we have to apply the quadratic equation formula and derive the answer.

Formula used: Quadratic equation formula:
 ${\text{x = }}\dfrac{{{{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$
To check the roots, we have to use = $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $

Complete step-by-step solution:
The given equation is ${{\text{x}}^2} = 5{\text{x + 10}}$ , we need to solve the equation using the quadratic equation formula.
Quadratic equation formula:
 ${\text{x = }}\dfrac{{{{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$
Rewriting the given equation, we get, ${\text{ - }}{{\text{x}}^2} + 5{\text{x + 10 = 0}}$
First of all, we have to check the roots. For that, we will use this formula $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $
Clearly in ${\text{ - }}{{\text{x}}^2} + 5{\text{x + 10 = 0}}$ ,
${\text{a = - 1}}$ ,
${\text{b = 5}}$ ,
${\text{c = 10}}$ .
Putting all the values in the formula, we get,
$\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} = \sqrt {{{\text{5}}^2}{\text{ - 4( - 1)(10)}}} $
Simplifying the equation,
$\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} = \sqrt {{\text{25 + 40}}} $
$\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} = \sqrt {65} $
Therefore, the roots exist and the roots are different and clear.
Now applying ${\text{ - }}{{\text{x}}^2} + 5{\text{x + 10 = 0}}$ in $\dfrac{{{{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$ ,
  ${\text{x = }}\dfrac{{ - 5 \pm \sqrt {{5^2} - 4( - 1)(10)} }}{{2( - 1)}}$
Simplifying the equation,
$\Rightarrow$${\text{x = }}\dfrac{{ - 5 \pm \sqrt {25 + 4(10)} }}{{ - 2}}$
$\Rightarrow$${\text{x = }}\dfrac{{ - 5 \pm \sqrt {25 + 40} }}{{ - 2}}$
Simplifying the numerator, we get,
$\Rightarrow$${\text{x = }}\dfrac{{ - 5 \pm \sqrt {65} }}{{ - 2}}$
Now we can expand the expression into two, as there is a $ \pm $ in the expression. One becomes plus and the other becomes minus.
$\Rightarrow$${\text{x = }}\dfrac{{ - 5 + \sqrt {65} }}{{ - 2}},\dfrac{{ - 5 - \sqrt {65} }}{{ - 2}}$
Taking minus out from the numerator, we can cancel the minus with the denominator
$\Rightarrow$${\text{x = }}\dfrac{{ - \left( {5 - \sqrt {65} } \right)}}{{ - 2}},\dfrac{{ - \left( {5 + \sqrt {65} } \right)}}{{ - 2}}$

Therefore, the roots are ${\text{x = }}\dfrac{{\left( {5 - \sqrt {65} } \right)}}{2},\dfrac{{\left( {5 + \sqrt {65} } \right)}}{2}$

Note: In these types of questions, students always miss an important step that will reduce their marks. 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 exists, check whether it is clear roots or equal roots. To check the roots, we have to find the value of $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $ first,
If $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $ is greater than $0$ , then the roots are different and clear.
If $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $ is equal to $0$, then the roots are equal.
If $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $ is less than $0$ , then the roots do not exist.