SOLUTION: If I have the quadratic function, f(x)= (x^2)-x-12, and I want to find the x-intercept, I have to set y = 0, and factor like so: 0 = (x-4)(x+3). The questions is: why can't I co

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: If I have the quadratic function, f(x)= (x^2)-x-12, and I want to find the x-intercept, I have to set y = 0, and factor like so: 0 = (x-4)(x+3). The questions is: why can't I co      Log On


   



Question 1009738: If I have the quadratic function, f(x)= (x^2)-x-12, and I want to find the x-intercept, I have to set y = 0, and factor like so: 0 = (x-4)(x+3).
The questions is: why can't I complete the square or factor like so: 0 = x(x-1)-12 (which leads to 0 = (x-1)(x-12))? Doing either of those alternative techniques results in different x-intercepts.

Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
You're not doing the algebra right. Completing
the square should work, though.
+x%5E2+-+x+-+12+=+0+
+x%5E2+-+x+=12+
+x%5E2+-+x+%2B+%28-1%2F2%29%5E2+=+12+%2B+%28-1%2F2%29%5E2+
+x%5E2+-+x+%2B+1%2F4+=+12+%2B+1%2F4+
+x%5E2+-+x+%2B+1%2F4+=+49%2F4+
+%28+x+-+1%2F2+%29%5E2+=+%287%2F2%29%5E2+
Take the square root of both sides
(1) +x+-+1%2F2+=+7%2F2+
and, also:
(2) +x+-+1%2F2+=+-7%2F2+
-------------------
(1) +x+=+8%2F2+
(1) +x+=+4+
and
(2) +x+=+-6%2F2+
(2) +x+=+-3+
------------------
So, the factors are:
+%28+x+-+4+%29%2A%28+x+%2B+3+%29+=+0+
You could also use the quadratic formula

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

If I have the quadratic function, f(x)= (x^2)-x-12, and I want to find the x-intercept, I have to set y = 0, and factor like so: 0 = (x-4)(x+3).
The questions is: why can't I complete the square or factor like so: 0 = x(x-1)-12 (which leads to 0 = (x-1)(x-12))? Doing either of those alternative techniques results in different x-intercepts.
You can use the "complete the square" method to solve for the x-intercepts/solutions to the equation/roots.
However, factoring f%28x%29+=+x%5E2+-+x+-+12 CANNOT be done the following way you proposed:
f%28x%29+=+x%5E2+-+x+-+12
0+=+x%28x+-+1%29+-+12
From this, you CANNOT obtain x - 1 and x - 12 as factors. That's not the proper way to factor a trinomial in
order to find its roots. The 1st method: 0+=+%28x+-+4%29%28x+%2B+3%29, though is indeed correct.