Solving quadratic equations by factoring method

Solve the equation below step-by-step by factoring method

$\large&space;\mathbf{6x^{2}-7x-20=0}$

Solving equations using factoring method is a daunting task as you have to carefully look for numbers that will produce roots of the quadratic equation.

Steps of factoring method

1. Put all terms in one side preferably the left side and put zero on one side preferably right side
2. Arrange the terms in the general form of quadratic equation
3. Find product of first and third term
4. Find two numbers or factors when multiplied they give the product in step 3. above and when added they give the second term
5. Rewrite the equation and substitute the middle term with the two factors you found in step 4.
6. Group and factorise the first two terms and last two terms
7. Find common linear factor, factorise further and completely
8. Equate each linear factor to zero
9. Obtain the roots

$\fn_jvn&space;\large&space;\mathbf{6x^{2}{\color{Magenta}-7x}-20=0}$

Find product of first and third term

$\inline&space;\fn_jvn&space;\large&space;\fn_jvn&space;\LARGE&space;\mathbf{\left&space;(&space;6x^{2}&space;\right&space;)\left&space;(&space;-20&space;\right&space;)=-120x^{2}}$

Find factors that give the product above; when added they give the sum of middle term

$\fn_jvn&space;\LARGE&space;\mathbf{\left&space;(&space;-15x\right&space;)\left&space;(&space;8x\right&space;)=-120x^{2}}$

Sum of factors that equal second term

$\fn_jvn&space;\large&space;\fn_jvn&space;\large&space;\mathbf{-15x+8x={\color{Magenta}&space;7x}}$

The factors summed up in this case are generated from testing various divisors of

$\fn_jvn&space;\LARGE&space;\mathbf{-120x^{2}}$

Let’s start

Let’s switch off from factoring method to completing the square method and see how beautifully the equation is solved.

$\large&space;\mathbf{6x^{2}-7x-20=0}$

This is a quadratic equation of the form

$\large&space;\\\\{\color{Magenta}&space;\mathbf{ax^{2}+bx+c=0}}\\$

Divide by 6 throughout to make the coefficient of x squared 1

$\large&space;\\\mathbf{x^{2}-\frac{7}{6}x-\frac{20}{6}=0}\\\\$

In order to complete the square, take half the coefficient of x in the second term, square the result and add to both sides

$\large&space;\mathbf{x^{2}-\frac{7}{6}x+\left&space;(&space;\frac{1}{2}\cdot-\frac{7}{6}&space;\\\\\right&space;)^{2}=\frac{20}{6}+\left&space;(&space;\frac{1}{2}\cdot-\frac{7}{6}&space;\right&space;)^{2}}$

To make LHS a perfect square express in the form of

$\fn_jvn&space;\large&space;\mathbf{{\color{Magenta}a^{2}-2ab+b^{2}=\left&space;(&space;a-b&space;\right&space;)^{2}}}$ $\fn_jvn&space;\large&space;\mathbf{x^{2}-\frac{7}{6}x+\left&space;(&space;\frac{1}{2}\cdot-\frac{7}{6}&space;\\\\\right&space;)^{2}=\mathbf{\left&space;(&space;x-\frac{7}{12}&space;\right&space;)^{2}}}$

On the RHS find the LCM of the fractions and add them together. In this case the LCM is 144

$\large&space;\\\mathbf{\left&space;(&space;x-\frac{7}{12}&space;\right&space;)^{2}=\frac{20}{6}+\frac{49}{144}}\\\\$

As you can see the RHS is simplified

$\large&space;\\\mathbf{\left&space;(&space;x-\frac{7}{12}&space;\right&space;)^{2}=\frac{529}{144}}\\\\$

Take the square roots on both sides

$\large&space;\mathbf{\sqrt{\left&space;(&space;x-\frac{7}{12}&space;\right&space;)^{2}}=\pm&space;\sqrt{\frac{529}{144}}}$

Taking square roots results in two solutions denoted by ±

$\large&space;\mathbf{x-\frac{7}{12}=\pm&space;\frac{23}{12}}$

Find values of x by adding the constant terms on the LHS to both sides

$\large&space;\mathbf{x=\frac{7}{12}\pm&space;\frac{23}{12}}$

Separate into two equations and simplify them by addition or subtraction

$\large&space;\mathbf{x=\frac{7}{6}+\frac{23}{12}}$ $\large&space;\mathbf{x=\frac{7}{6}-\frac{23}{12}}$

Find the LCM and get the results as shown below. Since the denominators are the same in both cases ( 6 and 12 ), the LCM is 12

Therefore,

$\large&space;\mathbf{x=\frac{30}{12}=\frac{5}{2}}$

Or

$\large&space;\mathbf{x=-\frac{16}{12}=-\frac{4}{3}}$

That is it!