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Thread: Find A and B

  1. #1
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    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?
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  2. #2
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    Re: Find A and B

    Quote Originally Posted by mathdad1965 View Post
    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

    $a=1$

    $b=A$

    $c=B$
    Thanks from mathdad1965
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  3. #3
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    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
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  4. #4
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    Re: Find A and B

    Quote Originally Posted by romsek View Post
    have you seen the quadratic formula?

    just apply that here with

    $a=1$

    $b=A$

    $c=B$
    I got it.
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  5. #5
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    Re: Find A and B

    Quote Originally Posted by Idea View Post
    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?
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  6. #6
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    Re: Find A and B

    Quote Originally Posted by mathdad1965 View Post
    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 $B=-2$.
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  7. #7
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    Re: Find A and B

    Quote Originally Posted by Plato View Post
    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.
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  8. #8
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    Re: Find A and B

    A=1 , B=-2 as shown above ( Plato #6)
    but also A=B=0
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  9. #9
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    Re: Find A and B

    Quote Originally Posted by mathdad1965 View Post
    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

    $r_1+r_2 = -A$

    $r_1 \times r_2 = B$
    Thanks from mathdad1965
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  10. #10
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    Re: Find A and B

    Quote Originally Posted by Idea View Post
    A=1 , B=-2 as shown above ( Plato #6)
    but also A=B=0
    If we allow $B=0$ then \forall A\in\mathbb{R} will work.
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  11. #11
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    Re: Find A and B

    Quote Originally Posted by romsek View Post
    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!
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  12. #12
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    Re: Find A and B

    If we allow $B=0 then$ \forall A\in\mathbb{R} will work.
    DISREGARD PLEASE. I missed the form x^2+Ax+B.
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  13. #13
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    Re: Find A and B

    I thank everyone for your input.
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  14. #14
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    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.
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  15. #15
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    Re: Find A and B

    Quote Originally Posted by mathdad1965 View Post
    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.
    No, it's not. "0= 0" is true for any values of A and B.

    Is the formula Ax + By + C = 0 used in any way here? Can someone get me started?
    If both A and B satisfy x^2+ Ax+ B= 0 then A^2+ A(A)+ B= 2A^2+ B= 0, so that B= -A^2, and B^2+ AB+ B= 0. Since B= -A^2, (-A^2)^2+ A(-A^2)+ (-A^2)= A^4- A^3- A^2= A^2(A^2- A- 1)= 0.
    Thanks from mathdad1965
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