Question

How do you solve \[{{x}^{2}}-2x-15=0?\]

Answer 1

Hint: We are given \[{{x}^{2}}-2x-15=0\] and to solve this we learn about the type of equation we are given and then learn the number of the solutions of the equation. We will learn how to factor the quadratic equation, we will use the middle term split to factor the term and we will simplify by taking common terms out. We also use the zero product rule to get our answer. To be sure about your answer we can also check by putting the acquired value of the solution in the given Equation and check whether they are the same or not.

Complete step by step answer:
We are given \[{{x}^{2}}-2x-15=0\] and we are asked to solve the given problem. First, we observe that it has a maximum power of ‘2’ so it is a quadratic equation. Now we should know that the quadratic equation has 2 solutions or we say an equation of power ‘n’ will have ‘n’ solution. Now as it is a quadratic equation we will change it into standard form \[a{{x}^{2}}+bx+c=0.\] As we look closely our problem is already in standard form
\[{{x}^{2}}-2x-15=0\]
Now we have to solve the equation \[{{x}^{2}}-2x-15=0.\] To solve this equation we first take the greatest common factor possibly available to the terms. As we can see that in \[{{x}^{2}}-2x-15=0,\] 1, – 2, – 15 has nothing in common. So, the equation remains the same.
\[{{x}^{2}}-2x-15=0\]
Now we will solve this using a method which is called by factoring. We factor using the middle term split. For the quadratic equation, \[a{{x}^{2}}+bx+c=0\] we will find such a pair of a term whose products is the same as \[a\times c\] and whose sum or difference will be equal to the b. Now in \[{{x}^{2}}-2x-15=0\] we have a = 1, b = – 2 and c = – 15. So,
\[a\times c=1\times -15=-15\]
We can see that \[-5\times 3=-15\] and their sum is – 5 + 3 = – 2. So, we use this to split.
\[{{x}^{2}}-2x-15=0\]
\[\Rightarrow {{x}^{2}}+\left( -5+3 \right)x-15=0\]
Opening the brackets, we get,
\[\Rightarrow {{x}^{2}}-5x+3x-15=0\]
Taking common in the first two and last two terms. We get,
\[\Rightarrow x\left( x-5 \right)+3\left( x-5 \right)=0\]
As (x – 5) is common, so simplifying further, we get,
\[\Rightarrow \left( x+3 \right)\left( x-5 \right)=0\]
Using the zero product rule which says if two terms product is zero that either one of them is zero. So, either
\[\Rightarrow x+3=0;x-5=0\]
So, we get,
\[\Rightarrow x=-3;x=5\]

Hence the solutions are x = – 3 and x = 5.

Note: We can also find the solution using another method called a quadratic formula. For equation \[a{{x}^{2}}+bx+c=0\] the solution is given as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.\] So for our equation \[{{x}^{2}}-2x-15=0\] we have a = 1, b = – 2 and c = – 15. So, we get,
\[\Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 1\times -15}}{2\left( 1 \right)}\]
On simplifying, we get,
\[\Rightarrow x=\dfrac{2\pm \sqrt{64}}{2}\]
As, \[\sqrt{64}=8,\] we get,
\[\Rightarrow x=\dfrac{2\pm 8}{2}\]
So, we get,
\[\Rightarrow x=\dfrac{10}{2};x=\dfrac{-6}{2}\]
On simplifying, we get,
\[\Rightarrow x=5;x=-3\]