# Thread: Find A and B

1. ## Find A and B

For which values of A and B will the roots of the equation x^2 + Ax + B = 0 be A and B?

The question is asking to find the value of A and B that will yield 0 = 0.

Is the formula Ax + By + C = 0 used in any way here? Can someone get me started?

2. ## Re: Find A and B

For which values of A and B will the roots of the equation x^2 + Ax + B = 0 be A and B?

The question is asking to find the value of A and B that will yield 0 = 0.

Is the formula Ax + By + C = 0 used in any way here? Can someone get me started?
have you seen the quadratic formula?

just apply that here with

$a=1$

$b=A$

$c=B$

3. ## Re: Find A and B

we could also look at
the sum of the roots = -A = A + B
and the product of the roots = B = A*B

4. ## Re: Find A and B

Originally Posted by romsek
have you seen the quadratic formula?

just apply that here with

$a=1$

$b=A$

$c=B$
I got it.

5. ## Re: Find A and B

Originally Posted by Idea
we could also look at
the sum of the roots = -A = A + B
and the product of the roots = B = A*B
How is it done this way?

6. ## Re: Find A and B

For which values of A and B will the roots of the equation x^2 + Ax + B = 0 be A and B?
If each of $A~\&~B$ is a root, then by the factor theorem we have:
$(x-A)(x-B)=x^2-(A+B)+AB=0$. Thus we need $A=-(A+B)~\&~AB=B$
From the second, we get $A=1$. Then that implies $B=-2$.

7. ## Re: Find A and B

Originally Posted by Plato
If each of $A~\&~B$ is a root, then by the factor theorem we have:
$(x-A)(x-B)=x^2-(A+B)+AB=0$. Thus we need $A=-(A+B)~\&~AB=B$
From the second, we get $A=1$. Then that implies $B=-2$.
Ok.

8. ## Re: Find A and B

$\displaystyle A=1 , B=-2$ as shown above ( Plato #6)
but also $\displaystyle A=B=0$

9. ## Re: Find A and B

How is it done this way?
by the quadratic formula the roots of $a x^2 + b x + c$ are

$r_{1,2} = \dfrac{-b \pm \sqrt{b^2 - 4 a c}}{2a}$

$r_1 + r_2 = \dfrac{-b + \sqrt{b^2 - 4 a c}}{2a}+ \dfrac{-b -\sqrt{b^2 - 4 a c}}{2a} = \dfrac {-2b}{2a} = -\dfrac b a$

$r_1 \times r_2 = \dfrac{b^2 -(\sqrt{b^2 - 4ac})^2}{4a^2} = \dfrac{b^2 - (b^2-4ac)}{4a^2} = \dfrac{4 a c}{4 a^2} = \dfrac {c}{a}$

with

$a=1,~b=A,~c=B$ this translates to

$r_1+r_2 = -A$

$r_1 \times r_2 = B$

10. ## Re: Find A and B

Originally Posted by Idea
$\displaystyle A=1 , B=-2$ as shown above ( Plato #6)
but also $\displaystyle A=B=0$
If we allow $B=0$ then $\displaystyle \forall A\in\mathbb{R}$ will work.

11. ## Re: Find A and B

Originally Posted by romsek
by the quadratic formula the roots of $a x^2 + b x + c$ are

$r_{1,2} = \dfrac{-b \pm \sqrt{b^2 - 4 a c}}{2a}$

$r_1 + r_2 = \dfrac{-b + \sqrt{b^2 - 4 a c}}{2a}+ \dfrac{-b -\sqrt{b^2 - 4 a c}}{2a} = \dfrac {-2b}{2a} = -\dfrac b a$

$r_1 \times r_2 = \dfrac{b^2 -(\sqrt{b^2 - 4ac})^2}{4a^2} = \dfrac{b^2 - (b^2-4ac)}{4a^2} = \dfrac{4 a c}{4 a^2} = \dfrac {c}{a}$

with

$a=1,~b=A,~c=B$ this translates to

$r_1+r_2 = -A$

$r_1 \times r_2 = B$
Great job!

12. ## Re: Find A and B

If we allow $B=0 then$ $\displaystyle \forall A\in\mathbb{R}$ will work.
DISREGARD PLEASE. I missed the form $\displaystyle x^2+Ax+B$.

13. ## Re: Find A and B

I thank everyone for your input.

14. ## Re: Find A and B

From now on, I will not post questions just for the pleasure of knowing what to do. If a question is above my level of math, it will not be posted.

15. ## Re: Find A and B

If both A and B satisfy $\displaystyle x^2+ Ax+ B= 0$ then $\displaystyle A^2+ A(A)+ B= 2A^2+ B= 0$, so that $\displaystyle B= -A^2$, and $\displaystyle B^2+ AB+ B= 0$. Since $\displaystyle B= -A^2$, $\displaystyle (-A^2)^2+ A(-A^2)+ (-A^2)= A^4- A^3- A^2= A^2(A^2- A- 1)= 0$.