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

Answer 1

Genesis Allen

First subtract #12# from both sides to get #4x^2-13x-12 = 0#, then use the quadratic formula to find:

#x = (13 +- 19)/8#

That is #x=4# or #x=-3/4#

#4x^2-13x-12# is in the form #ax^2+bx+c# with #a=4#, #b=-13# and #c=-12#

It discriminant is given by the formula:

#Delta = b^2-4ac = 13^2 - (4xx4xx-12) = 169+192 = 361 = 19^2#

Since #Delta# is positive and a perfect square, the roots of our quadratic are rational, given by the formula:

#x = (-b+-sqrt(Delta))/(2a) = (13+-19)/8#

That is #x=32/8=4# or #x=-6/8 = -3/4#

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Answer 2

Jameson Allen

To solve the equation 4x^2 - 13x = 12, you can follow these steps:

Rearrange the equation to set it equal to zero: 4x^2 - 13x - 12 = 0.
Factor the quadratic expression: (4x + 3)(x - 4) = 0.
Set each factor equal to zero and solve for x: 4x + 3 = 0 or x - 4 = 0.
Solve for x in each equation: For 4x + 3 = 0: 4x = -3, x = -3/4. For x - 4 = 0: x = 4.

Therefore, the solutions are x = -3/4 and x = 4.

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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.