What an I doing wrong? I'm trying to get the original equation into parenthesis

2. Re: Basic quadratic formula issue

$(x+1)(x+\frac{1}{4})=0$ is the same equation as $(x+1)(4x+1)=0$

3. Re: Basic quadratic formula issue

Then why doesn't this problem work?

4. Re: Basic quadratic formula issue

Shakarri multiplied by 4 on both sides of $(x+ 1)(x+ \frac{1}{4})= 0$, then took the "4" on the left into the second parentheses, using the distributive law.

5. Re: Basic quadratic formula issue

But the book for this problem is getting a different answer somehow?

6. Re: Basic quadratic formula issue

Then tell us what answer the textbook is getting!

7. Re: Basic quadratic formula issue

Originally Posted by HallsofIvy
Then tell us what answer the textbook is getting!
Please check my previous post with the image

8. Re: Basic quadratic formula issue

Multiply out your brackets and get rid of any fractions, see what you end up with.

9. Re: Basic quadratic formula issue

The original equation was $4x^2+ 5x+ 1= 0$ and you used the quadratic formula to find x= -1 and x= -1/4 as solutions so that the factors are $(x+1)(x+ 1/4)$.
Now there are two ways to go:
1) You can multiply those factors to get $x^2+ x+ (1/4)x+ 1/4= x^2+ (5/4)x+ 1/4= 0$. You can then multiply both sides of that equation by 4 to get $4x^2+ 5x+ 1= 0$.

2) As Shakarri said, you can immediately multiply both sides of $(x+ 1)(x+ 1/4)= 0$ by 4 to get $(x+ 1)(4x+ 1)= 0$. Then multiply that to get $4x^2+ 4x+ x+ 1= 4x^2+ 5x+ 1= 0$.