For the solution of finding zeros flr -6x^2-13x+5
When i use factoring and completing the squares the zeros are 1/3 and -5/2
However when i used the quadratic formula it gave me -13+ - sqrt of 17/ 12
Why is that. Is it still tge same
How
[-b±√(b^2-4ac)]/2a
= [13±√(169+120)]/-12
= (13±√289)/-12
= (13±17)/-12
= 1/3 or -5/2
Did you take √ twice?
Let's break down the steps for finding the zeros of the quadratic equation -6x^2 - 13x + 5 using different methods.
1. Factoring method:
To find the zeros by factoring, we need to express the equation in the form (ax + b)(cx + d) = 0, where a, b, c, and d are constants.
For the equation -6x^2 - 13x + 5, we need to find two numbers that add up to -13 and multiply to -6 * 5 = -30. The numbers are -3 and -10.
So we rewrite the equation as:
-6x^2 - 13x + 5 = (-3x + 1)(2x - 5) = 0
Setting each factor equal to zero, we get:
-3x + 1 = 0 and 2x - 5 = 0
Solving for x in each equation, we have:
-3x = -1 => x = 1/3
2x = 5 => x = 5/2
So the zeros of the equation are x = 1/3 and x = 5/2.
2. Quadratic formula method:
The quadratic formula is x = (-b ± √(b^2 - 4ac))/(2a).
For the equation -6x^2 - 13x + 5, the coefficients are a = -6, b = -13, and c = 5.
Plugging these values into the quadratic formula, we have:
x = [ -(-13) ± √((-13)^2 - 4(-6)(5)) ] / (2(-6))
x = (13 ± √(169 + 120)) / (-12)
x = (13 ± √289) / (-12)
x = (13 ± 17) / (-12)
Simplifying further, we get:
x = (-13 + 17) / (-12) => x = 4 / (-12) => x = -1/3
and
x = (-13 - 17) / (-12) => x = -30 / (-12) => x = 5/2
So with the quadratic formula, we find the zeros x = -1/3 and x = 5/2.
In conclusion, both factoring and using the quadratic formula have provided the same zeros for the quadratic equation -6x^2 - 13x + 5, which are x = 1/3 and x = 5/2.