SOLUTION: Find the real solutions, if any, of the given equation. (3/4)x^2 - (1/4)x - (1/2) = 0 Let me see. I can use the quadratic formula. In the formula, a = 3/4, b =

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find the real solutions, if any, of the given equation. (3/4)x^2 - (1/4)x - (1/2) = 0 Let me see. I can use the quadratic formula. In the formula, a = 3/4, b =       Log On


   

Question 1207404: Find the real solutions, if any, of the given equation.


(3/4)x^2 - (1/4)x - (1/2) = 0


Let me see.

I can use the quadratic formula. In the formula, a = 3/4, b = -1/4 and c = -1/2.

Yes?
Found 3 solutions by math_tutor2020, ikleyn, MathTherapy:
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

You are correct.

Here's part of the scratch work involving the discriminant.
d = b^2 - 4ac
d = (-1/4)^2 - 4*(3/4)*(-1/2)
d = 1/16 + 3/2
d = 1/16 + 24/16
d = (1+24)/16
d = 25/16
I'll let you finish up.

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.

The short answer to your question is "Yes".



But there is another way, much more attractive.


Multiply all the terms of your equation by 4.


You will get then other, EQUIVALENT quadratic equation

    3x^2 - x - 2 = 0.


Apply now the quadratic formula, which you know

    x%5B1%2C2%5D = %281+%2B-+sqrt%28%28-1%29%5E2+-+4%2A3%2A%28-2%29%29%29%2F%282%2A3%29 = %281+%2B-+sqrt%2825%29%29%2F6.


Thus  x%5B1%5D = %281+%2B+sqrt%2825%29%29%2F6 = %281+%2B+5%29%2F6 = 1

and   x%5B2%5D = %281+-+sqrt%2825%29%29%2F6 = %281+-+5%29%2F6 = -4%2F6 = -2%2F3.


So, the solutions are the numbers  1  and   -2%2F3.

Solved.

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On solving quadratic equations using the quadratic formula,  see the lessons
    - Introduction into Quadratic Equations
    - PROOF of quadratic formula by completing the square
in this site.




Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!
Find the real solutions, if any, of the given equation. 

(3/4)x^2 - (1/4)x - (1/2) = 0

Let me see.

I can use the quadratic formula. In the formula, a = 3/4, b = -1/4 and c = -1/2.

Yes? 

You could but it'll be a bit tedious, with all those fractions. It's much easier to get rid of the fractions by 
multiplying the equation by its LCD, 4. This'll give you: 

Now, you can use the quadratic equation formula, but this quadratic (as is obvious, MAYBE), can be factored by
using the "ac" method, which requires 2 factors that have a PRODUCT of - 6 (a * c = + 3 * - 2), and a SUM of - 1
(b, or the coefficient on "x"). These 2 FACTORS are - 3 and + 2. 
We then get the following: 3x2 - x - 2 = 0 
                     3x2 - 3x + 2x - 2 = 0 ----- Replacing - x with - 3x + 2x
                     3x2 - 3x + 2x - 2 = 0 ----- PAIRing binomials
                  3x(x - 1) + 2(x - 1) = 0
                       (3x + 2)(x - 1) = 0 ----- FACTORED form
              3x + 2 = 0       | x - 1 = 0
                  3x = - 2     |     x = 0 + 1
                  highlight_green%28matrix%281%2C3%2C+x%2C+%22=%22%2C+highlight%28-+2%2F3%29%29%29   OR   highlight_green%28matrix%281%2C3%2C+x%2C+%22=%22%2C+highlight%281%29%29%29