How do you solve #4x^2=13x+12#?

Question

Hazel Anglin · Accepted Answer

#(4x +3)(x-4)#

Find the factors of 4 and 12 which subtract to give 13.

However, 13 is an odd number which can only happen from the difference between an odd and an even number. This is a clue that the factors of 4 are not 2 and 2, because any product of these even numbers will give even numbers. Similarly the factors of 12 are not 2 and 6, because they are both even. We find the following cross products and subtract them:

#4" "3 rArr 1 xx 3 = 3# #1" 4" rArr 4xx4 = 16" "16-3 =13#

However, we need -13, so the combination must be -16 and +3.

#(4x +3)(x-4)#, so the

Eva Abramson · Answer

undefined To solve the equation $4x^2 = 13x + 12$, follow these steps:

1. Rewrite the equation in the standard form by moving all terms to one side: $4x^2 - 13x - 12 = 0$.
2. Factor the quadratic expression, if possible. If not, use the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a = 4$, $b = -13$, and $c = -12$.
3. Substitute the values of $a$, $b$, and $c$ into the quadratic formula and simplify.
4. Solve for $x$ by finding the two possible values using the plus and minus signs in the formula.

The solutions to the equation $4x^2 = 13x + 12$ are the values of $x$ obtained from the quadratic formula.