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How do you simplify this equation?: 2x^2-9x-5/ 4x^2-1
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We note that the part below ratio can be write as `4(x+1/2)(x-1/2)` The part above ratio as: `2(x+1/2)(x-5)`
So: `(2x^2-9x-5)/(4x^2-1)= (2(x+1/2)(x-5))/(4(x+1/2)(x-1/2))=` `(x-5)/(2(x-1/2))` `=(x-5)/(2x-1)`
Posted by oldnick on May 9, 2013 at 12:52 AM (Answer #1)
High School Teacher
We can use Factoring here.
For the top, we can apply the ac method.
Identify first what is a, b, and c.
a = 2, b = -9 and c = -5. Take note that the top is in the form ax^2 + bx + c.
Multiply a and c. 2 *-5 = -10.
We will find two numbers which when we multiply we get - 10 (value of ac), and when we add we get -9 (value of b).
Let us lists the pair factors of -10 and their sum:
-10 and 1 : - 10 + 1 = -9
10 and - 1 ; 10 + (-1) = 9
-5 and 2 ; -5 + 2 = -3
5 and - 2 ; 5 + (-2) = 3
Therefore the two numbers we are looking for are -10 and 1.
We will rewrite the middle terms using the numbers we got.
`2x^2 -10x + 1x - 5`
Split the terms into two groups, where each group has a gcf.
`(2x^2 - 10x) + (1x - 5)`
Factor out the gcf from each group.
`2x(x - 5) + 1(x - 5)`
Factor out the common factor.
`(x - 5)(2x + 1)`
For the bottom, we will use the Difference of Two Squares formula.
`a^2 - b^2 = (a - b)(a + b)`
`4x^2 - 1 = (2x)^2 - (1)^2 = (2x - 1)(2x + 1)`
So, we will have:
`((x - 5)(2x + 1))/((2x - 1)(2x+1))`
Cancel the common factor on top and bottom.
`(x - 5)/(2x - 1)`
Posted by violy on May 9, 2013 at 2:16 AM (Answer #2)
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