Question

How do you solve \[5{{x}^{2}}-7x-6=0\] by factoring?

Answer 1

Hint: We are given \[5{{x}^{2}}-7x-6=0\] and we have to solve this. We will learn about the type of equation we are given and then learn the number of the solutions of the equation. We will also 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 in the given equation and check whether they are the same or not.

Complete step by step answer:
We are given \[5{{x}^{2}}-7x-6=0\] 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 an ‘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 \[5{{x}^{2}}-7x-6=0.\]
Now, we have to solve the equation \[5{{x}^{2}}-7x-6=0.\] To solve this equation we first take the greatest common factor possibly available to the terms. As we can see that in \[5{{x}^{2}}-7x-6=0,\] 5, – 7 and – 6 have nothing in common. So, the equation remains the same.
\[5{{x}^{2}}-7x-6=0\]
Now to solve our problem we will use the Middle term split method. For quadratic equations \[a{{x}^{2}}+bx+c=0\] we will find such a pair of terms whose products are the same as \[a\times c\] and whose sum difference will be equal to b. Now in \[5{{x}^{2}}-7x-6=0\] we have a = 5, b = – 7 and c = – 6. So,
\[a\times c=5\times \left( -6 \right)=-30\]
We can see that \[-10\times 3=30\] and their sum is – 10 + 3 = – 7. So, we use this to split.
\[5{{x}^{2}}-7x-6=0\]
\[\Rightarrow 5{{x}^{2}}+\left( -10+3 \right)x-6=0\]
Taking common in first two and last two terms we get,
\[\Rightarrow 5x\left( x-2 \right)+3\left( x-2 \right)=0\]
As x – 2 is common, so simplifying further,
\[\Rightarrow \left( x-2 \right)\left( 5x+3 \right)=0\]
Using zero product rule which says if two terms product is zero that either one of them is zero. So either
\[\Rightarrow x-2=0;5x+3=0\]
\[\Rightarrow x=2;x=\dfrac{-3}{5}\]
Hence the solutions are x = 2 and \[x=\dfrac{-3}{5}.\]

Note:
We can also find the solution using the quadratic formula. By the method of quadratic formula, the solution will be found easily. 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 \[5{{x}^{2}}-7x-6=0,\] we have a = 5, b = – 7 and c = – 6. So, using these above we get,
\[\Rightarrow x=\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\times 5\times \left( -6 \right)}}{2\left( 5 \right)}\]
On simplifying, we get,
\[\Rightarrow x=\dfrac{7\pm \sqrt{169}}{10}\]
\[\Rightarrow x=\dfrac{7\pm 13}{10}\]
So, the solutions are
\[\Rightarrow x=\dfrac{7+13}{10};x=\dfrac{7-13}{10}\]
On simplifying, we get,
\[\Rightarrow x=2;x=\dfrac{-3}{5}\]