Question

Solve $x^2+5x-50=0$.

Answer 1

Hint: In order to determine the factors of the above quadratic equation use the Splitting up the middle term method or also called mid-term factorization method. In which the middle term would be splitted into two parts and similar values will be taken common and equated to zero to obtain the roots.

Complete step by step solution:
Since, the highest degree in the given polynomial is two, so it’s a quadratic equation.
Given a quadratic equation $x^2+5x-50$ ,let it be $f(x)$
$f(x)=x^2+5x-50$
Comparing the equation with the standard Quadratic equation $a{x}^2+bx+c$
$a$ becomes $1$
$b$ becomes $5$
And $c$ becomes $-50$
To find the quadratic factorization we’ll use splitting up the middle term method
So first calculate the product of coefficient of $x^2$ and the constant term which comes to be
$$=1\times \left(-50\right)=\left(-50\right)$$
Now second Step is to the find the $2$ factors of the number $-50$ such that whether, addition or subtraction of those numbers is equal to the middle term or coefficient of $x$ and the product of those factors results in the value of constant.
So, if we factorize $-50$ , the answer comes to be $10$ and $-5$ as $10-5=5$ that is the middle term. and $10\times \left(-5\right)=\left(-50\right)$ which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained, so equation $f(x)$ becomes
$f(x)=x^2+10x-5x-50=0$
Now taking common from the first $2$ terms and last $2$ terms
$f(x)=x\left(x+10\right)-5\left(x+10\right)=0$
Finding the common binomial parenthesis, the equation becomes
$f(x)=(x-5)(x+10)=0$
Hence, we have successfully factorized our quadratic equation.
Therefore, the factors are $\left(x-5\right)$ and $\left(x+10\right)$.
Equating the factors with zero to get the roots:
 $$\begin{gathered} \left(x-5\right)=0 \\ x=5 \end{gathered}$$
And,
 $$\begin{gathered} \left(x+10\right)=0 \\ x=\left(-10\right) \end{gathered}$$

Therefore, the roots of the quadratic equation $x^2+5x-50=0$ are: $x=5$ and $x=\left(-10\right)$.

Note:
Alternative Approach:
You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x_1={\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}}$ and $x_2={\displaystyle \frac{-b-\sqrt{b^2-4ac}}{2a}}$
$x_1$, $x_2$ are root to quadratic equation $a{x}^2+bx+c$
For the equation given above $x^2+5x-50=0$ , comparing the equation with the standard Quadratic equation $a{x}^2+bx+c$
$a$ becomes $1$
$b$ becomes $5$
And $c$ becomes $-50$
Substituting these values in $x_1={\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}}$ and $x_2={\displaystyle \frac{-b-\sqrt{b^2-4ac}}{2a}}$ , and we get:
For first root:
$$\begin{gathered} x_1={\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}} \\ \Rightarrow x_1={\displaystyle \frac{-5+\sqrt{{\left(-5\right)}^2-4\left(1\right)\left(-50\right)}}{2\left(1\right)}} \\ \Rightarrow x_1={\displaystyle \frac{-5+\sqrt{25+200}}{2}} \\ \Rightarrow x_1={\displaystyle \frac{-5+\sqrt{225}}{2}} \end{gathered}$$

Since, we know that $\sqrt{225}=15$ , so opening the parenthesis above:
$$\begin{gathered} x_1={\displaystyle \frac{-5+\sqrt{225}}{2}} \\ \Rightarrow x_1={\displaystyle \frac{-5+15}{2}}={\displaystyle \frac{10}{2}}=5 \end{gathered}$$
Therefore, the first root is $x_1=5$.

Similarly, for second root:
$$\begin{gathered} x_2={\displaystyle \frac{-b-\sqrt{b^2-4ac}}{2a}} \\ \Rightarrow x_2={\displaystyle \frac{-5-\sqrt{{\left(-5\right)}^2-4\left(1\right)\left(-50\right)}}{2\left(1\right)}} \\ \Rightarrow x_2={\displaystyle \frac{-5-\sqrt{25+200}}{2}} \\ \Rightarrow x_2={\displaystyle \frac{-5-\sqrt{225}}{2}} \end{gathered}$$

Since, we know that $\sqrt{225}=15$ , so opening the parenthesis above:
$$\begin{gathered} x_2={\displaystyle \frac{-5-\sqrt{225}}{2}} \\ \Rightarrow x_2={\displaystyle \frac{-5-15}{2}}={\displaystyle \frac{-20}{2}}=\left(-10\right) \end{gathered}$$
Therefore, the first root is $x_2=\left(-10\right)$.
Hence, the factors will be $\left(x-5\right)and\left(x+10\right)$.

1. Mid-term factorization and Quadratic formulas are always applied on Quadratic Equations.
2. Any one of the above methods can be used to find the roots of the quadratic equation.
3. We can leave the values till factors until it’s given to find the roots.
4. Don’t forget to compare the given quadratic equation with the standard one every time.