How would you solve for "x" in: 40=3x+x^2

Question

How would I solve for x in the short equation of 40=3x+x^2

Answer 1

Hi AJ,

Here is another suggestion for solving the problem using these steps:

40 = 3x + x²

40-40 = 3x + x² - 40  <-- Move all the terms to 1 side of the equal sign, so the equaton can be set to the value of ZERO and written as ax² + bx + c, where a, b  are coefficients and c is a constant.

0 = 3x + x² - 40 <-- This is not the proper way to display a quadratic equation, so we rewrite it as:

x² + 3x - 40 = 0

One way to solve a quadratic equations that follows the standard form of ax² + bx + c is to find a pair of factors of c that when ADDED together will equal b.

The factors of 40 are: 1, 2, 4, 5, 8, 10, 20 and 40.  A quick glance and we find that 8 - 5 = 3, so our factors must be either 5, -5, 8, or -8.

Since the coefficient on c is NEGATIVE, we have to choose one positive factor and one negative factor...REMEMBER a POS times and NEG is NEG.

5+(-8) = -3

8+ (-5) = 3

So the values to be used in the 2 factors for the quadratic equation are 8 and -5.  This can be proven by writing out the factors and multiplying the binomials

(x + 8)(x + (-5)) => (x + 8)(x - 5) => x2 + (-5x) + 8x + (-40) => x² + (8x - 5x) - 40 => x² + 3x - 40

We got back to the original equation so our factors for he quadratic equation are correct.

Don't forget to use the ZPP (Zero Product Property) to solve for the actual values of X...if a*b = 0, then either a=0 or b=0, so you can set each factor equal to 0. e.g. (x + 1) = 0 ==> x = -1.

NOTE:  Not all quadratic equations can be solved this way and when this method does not work we generally turn to the quadratic formula to solve quadratic equations.

Answer 2

40 = 3x + x²

rearrange the equation

x² + 3x  - 40 = 0

factor

(x + 8) * (x - 5) = 0

so, x = -8 or 5