roots of the quadratic

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Richard B

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The roots of the quadratic [MATH]x^2+8x+4[/MATH] are the same as the roots of the quadratic [MATH]Ax^2+Bx+1[/MATH]. What is [MATH]A+B[/MATH]?

I tried using the quadratic formula on both quadratics,


[MATH]x=\frac{-b\pm \sqrt{b^2-4a}}{2a}[/MATH]
[MATH]x=-4\pm2\sqrt{3}[/MATH]

then setting then equal to each other.

[MATH]\frac{-b\pm \sqrt{b^2-4a}}{2a}=-4\pm2\sqrt{3}[/MATH]
but when I try to solve for [MATH]a[/MATH] and [MATH]b[/MATH] it doesn't work.
 
Please show your work, so we can see what went wrong.

But I'd start differently, by thinking about how you can change a quadratic function without changing its zeros (roots, as you are calling them). Think about what the factored form implies, for example.
 
Using your method, we obtain the system:

[MATH]B=8A[/MATH]
[MATH]\frac{B^2-4A}{4A^2}=12[/MATH]
Substituting for \(B\) from the first into the second, what do we get?
 
Richard B said:
The roots of the quadratic [MATH]x^2+8x+4[/MATH] are the same as the roots of the quadratic [MATH]Ax^2+Bx+1[/MATH]. What is [MATH]A+B[/MATH]?

I tried using the quadratic formula on both quadratics,


[MATH]x=\frac{-b\pm \sqrt{b^2-4a}}{2a}[/MATH]
[MATH]x=-4\pm2\sqrt{3}[/MATH]

then setting then equal to each other.

[MATH]\frac{-b\pm \sqrt{b^2-4a}}{2a}=-4\pm2\sqrt{3}[/MATH]
but when I try to solve for [MATH]a[/MATH] and [MATH]b[/MATH] it doesn't work.
Click to expand...
Possibly you made a mistake in your work. If you had shown us your work one of the helpers on the forum could have pointed it out to you.
You thought of doing it a particular way. I think that we should proceed in that direction.

First I want to say that A and a are NOT the same. Same with B and b.

I think that now you should say that -B/(2A) = -4. Or B=8A.

Note that 2sqrt(3) = sqrt(12). Then we have 12 = (B^2-4A)/(4A^2). Substituting 8A for B we get 12 = (64A^2 - 4A)/(4A^2) = 16 - 1/A. So 1/A =4. Then A = 1/4. B=8*(1/4) = 2. A+B = 2 1/4
 
Here is the cleanest way to work this problem.

You need to realize that if you multiply a quadratic by any non-zero number the zeros will still be the same.

So I would change x^2 + 8x + 4 by dividing by 4 to get the constant to be 1! This gives me .25X^2 + 2X + 1. A=.25, B=2 and A+B = 2.25 !!!
 
MarkFL said:
Using your method, we obtain the system:

[MATH]B=8A[/MATH]
[MATH]\frac{B^2-4A}{4A^2}=12[/MATH]
Substituting for \(B\) from the first into the second, what do we get?
Click to expand...
Mark, I did not read your post very carefully as I went beyond your hint. Sorry!
 
MarkFL said:
Yes, this is what I get too. How about if we were to write:

[MATH]Ax^2+Bx+1=k(x^2+8x+4)[/MATH]
Click to expand...
I got the same thing!

[MATH]Ax^2+Bx+1=kx^2+8kx+4k[/MATH]
[MATH]B=8k[/MATH][MATH]A=k[/MATH][MATH]1=4k[/MATH]
[MATH]k=\frac14[/MATH]
[MATH]B=2[/MATH]
[MATH]A=\frac14[/MATH]
I put in the answer and it was correct, thanks for the help
 
Richard B said:
I got the same thing!

[MATH]Ax^2+Bx+1=kx^2+8kx+4k[/MATH]
[MATH]B=8k[/MATH][MATH]A=k[/MATH][MATH]1=4k[/MATH]
[MATH]k=\frac14[/MATH]
[MATH]B=2[/MATH]
[MATH]A=\frac14[/MATH]
I put in the answer and it was correct, thanks for the help
Click to expand...
Did you try to divide x^2 + 8x + 4 by 4 and then equate the coefficient?
 
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