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Thread: Finding the values of a and b

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Question Finding the values of a and b

    Hello everyone. This question is apparently unsolvable:

    If $\displaystyle x = 3$ or $\displaystyle -4$ are the solutions of the equation $\displaystyle x^2+ax+b=0$, find the values of $\displaystyle a$ and $\displaystyle b$.

    The keyword in this irksome question would be the word 'or'. So it denotes that that this involves quadratic formulas.

    Can anyone give me a clue so that I may make a breakthrough in understanding this problem? Thank you so much!
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  2. #2
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    The wording is a bit iffy, but they actually meant that the solutions for that equation is x = 3 AND x = -4.
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  3. #3
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    Quote Originally Posted by PythagorasNeophyte View Post
    This question is apparently unsolvable:

    If $\displaystyle x = 3$ or $\displaystyle -4$ are the solutions of the equation $\displaystyle x^2+ax+b=0$, find the values of $\displaystyle a$ and $\displaystyle b$.
    No this question IS solvable. And quite easily I might add.

    We know the quadratic formula as having a $\displaystyle \pm$ which yields 2 answers.

    Substituting in values from $\displaystyle x^2+ax+b=0$ into the quadratic formula we get:

    $\displaystyle x=\dfrac{-a + \sqrt{a^2 - 4 \times 1 \times b}}{2}$ and $\displaystyle x=\dfrac{-a - \sqrt{a^2 - 4 \times 1 \times b}}{2}$

    We know that the minus squareroot usually gives us the smaller answer of x. So then we just substitute in our x answers to get:

    $\displaystyle 3=\dfrac{-a + \sqrt{a^2 - 4 \times b}}{2}$ and $\displaystyle -4=\dfrac{-a - \sqrt{a^2 - 4 \times b}}{2}$

    Now solve for a and b using simultaneous equation.
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  4. #4
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    Quote Originally Posted by Educated View Post
    No this question IS solvable. And quite easily I might add.

    We know the quadratic formula as having a $\displaystyle \pm$ which yields 2 answers.

    Substituting in values from $\displaystyle x^2+ax+b=0$ into the quadratic formula we get:

    $\displaystyle x=\dfrac{-a + \sqrt{a^2 - 4 \times 1 \times b}}{2}$ and $\displaystyle x=\dfrac{-a - \sqrt{a^2 - 4 \times 1 \times b}}{2}$

    We know that the minus squareroot usually gives us the smaller answer of x. So then we just substitute in our x answers to get:

    $\displaystyle 3=\dfrac{-a + \sqrt{a^2 - 4 \times b}}{2}$ and $\displaystyle -4=\dfrac{-a - \sqrt{a^2 - 4 \times b}}{2}$

    Now solve for a and b using simultaneous equation.
    That is a very methodical method, great for understanding concepts.

    If you wish to know, a quicker way is to simply expand (x-3)(x+4) = 0.
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  5. #5
    Senior Member Educated's Avatar
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    I guess I overcomplicated things...
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  6. #6
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    Another way: If x= 3 and x= -4 are roots of the equation $\displaystyle x^2+ ax+ b$, then [tex](x- 3)(x+ 4)= x^2+ ax+ b[tex]. Just multiply out the left side to find a and b.

    I see that Gusbob already said that.
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