Home > Algebra > Comparing Methods for Solving Quadratics

How do you solve #x^2-x=12#?

Answer 1

Abigail Allen

We can use the Sum-Product method.

Bring everything to one side: #x^2-x=12->x^2-x-12=0#

We now have a quadratic equation of the form #ax^2+bx+c=0# where #a=1, b=-1 and c=-12#

We find two numbers that will give #c=-12# as a product and #b=-1# as a sum (or difference). We can try #1*12, 2*6,3*4#

#3and4# will fit, with a #-#sign to the #4#, as that would make the sum #3-4=-1# and the product #3*-4=-12#

Now we can rewrite the equation: #(x+3)(x-4)=0#

Which leaves us with two possibilities: #(x+3)=0->x=-3# OR #(x-4)=0->x=4#

Answer: #x=-3 or x=4#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Jameson Allen

To solve the equation (x^2 - x = 12), follow these steps:

Rearrange the equation to bring all terms to one side to set it equal to zero: (x^2 - x - 12 = 0).
Factor the quadratic equation: ((x - 4)(x + 3) = 0).
Apply the zero-product property: (x - 4 = 0) or (x + 3 = 0).
Solve for (x): (x = 4) or (x = -3).

Therefore, the solutions to the equation are (x = 4) and (x = -3).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.