How do you solve #4x^2 = 12x + 40#?

2 Answers
Aug 1, 2015

Answer:

#x_(1,2) = (3 +-7)/2#

Explanation:

You could solve this quadratic equation by using the quadratic formula.

For a quadratic equation that has the general form

#color(blue)(ax^2 + bx + c = 0)#

the quadratic formula takes the form

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)#

So, in order to be able to use this formula, get your quadratic to the general form by adding #(-12x - 40# to both sides of the equation

#4x^2 - 12x - 40 = color(red)(cancel(color(black)(12x))) + color(red)(cancel(color(black)(40))) - color(red)(cancel(color(black)(12x))) - color(red)(cancel(color(black)(40)))#

#4x^2 - 12x - 40 = 0#

You can simplify this by dividing all the terms by #4# to get

#(color(red)(cancel(color(black)(4)))x^2)/color(red)(cancel(color(black)(4))) - 12/4x - 40/4 = 0#

#x^2 - 3x - 10 = 0#

The two solutions of the equation will thus be

#x_(1,2) = (-(-3) +- sqrt( (-3)^2 - 4 * 1 * (-10)))/(2 * 1)#

#x_(1,2) = (3 +- sqrt(49))/2 = (3 +- 7)/2 = {(x_1 = (3 + 7)/2 = color(green)(5)), (x_2 = (3 - 7)/2 = color(green)(-2)) :}#

Aug 1, 2015

Answer:

Solve #y = 4x^2 - 12x - 40 = 0#

Ans: -2 and 5.

Explanation:

#y = 4(x^2 - 3x - 10) = 0#
I use the new Transforming Method. Find 2 numbers knowing sum (-3) and product (-10). Roots have opposite signs.
Factor pairs of (-10) --> (-1, 10)(-2, 5). This sum is 3 = -b.
Then, the 2 real roots are: -2 and 5.