Question

Decide equation using only factors to solve this problem
\(12x^{2}\ +\ 5x\ -\ 2 = 0\)

Answer 1

In order to factor the quadratic, we need to find the values of a, b and c for the given quadratic \(12x^{2}\ +\ 5x\ -\ 2 = 0\) on comparing with standard quadratic
\(ax^{2}\ +\ bx\ +\ c = 0\)
\(12x^{2}\ +\ 5x\ -\ 2 = 0\)
\(a = 12\, b = 5,\ c =\ -2.\)
Now, let us apply ac method to factor the quadratic.
We got \(ac=\ -24\ and\ b=5.\) So, we need to find the factors of -24, those add upto 5.
\(ac = 12 \cdot\ -2 =\ -24\)(Product)
\(b = 5\)(Sum)
-24 could be factored in two numbers 8 and -3 that add upto 5.
Therefore, middle term 5x of expression \(12x^{2}\ +\ 5x\ -\ 2 =0,\) could be break into two terms 8x and -3x.
Then we need to make the break the expression into groups.
Factor out 3x form \(12x^{2}\ -\ 3x\) and factor out 2 from:
\(8x\ -\ 2.\)
Then factor out common parenthesis \((4x\ -\ 1).\)
We got factored form \((4x\ -\ 1) (3x\ +\ 2) = 0\)
\(12x^{2}\ +\ 5x\ -\ 2 = 0\)
\((12x^{2}\ -\ 3x)\ +\ (8x\ -\ 2) = 0\)
\(3x(4x\ -\ 1)\ +\ 2 (4x\ -\ 1) = 0\)
Factor out common parenthesis \((4x\ -\ 1)\)
\((4x\ -\ 1)(3x\ +\ 2) = 0\)
By applying zero product rule, we need to set each factor equal to zero and solve for x.
We got \(x = \frac{1}{4},\ \frac{-2}{3}.\)
\(4x\ -\ 1 = 0\ or\ 3x\ +\ 2 = 0\)
\(+1 +1 -2 -2\)
\(\frac{4x}{4} = \frac{1}{4}\ or\ \frac{3x}{3} = \frac{-2}{3}\)
\(x = 1.4\ or\ x = \frac{-2}{3}\)