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Using the Principle of Zero Products, create a quadratic in standard form that has roots of $-1$ and $\frac{3}{2}$. Values of $a, b$, and $c$ in your final form should all be elements of the set of integers.

I really don't understand how to do this. I tried to use $(x + \frac{1}{4})2 = \frac{5}{16}$, to create a quadratic, but I can't quite seem to figure it out.

Jessie
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RetroBreads
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    I don't understand what your attempt has to do with the problem. Where do $\frac14$ and $\frac5{16}$ come from? The point is that $a$ is a root, if and only if $x-a$ is a factor. – saulspatz Feb 23 '21 at 18:15
  • You want to find $ax^2 + bx + c$ so that 1) $\frac 12$ and $1$ are solutions and 2) that $a,b,c$ are all integers. $(x - \frac 12)(x -(-1))$ will be a quadratic with those solutions, but it might not have integer coefficients. To get integer coefficients we can multiply it by some constant. ... so what $ax^2 + bx + c$ has $\frac 12$ and $-1$ as roots? – fleablood Feb 23 '21 at 18:44

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You want to find $ax^2 + bx + c$ so that

  1. $\frac 12$ and $1$ are solutions and
  2. that $a,b,c$ are all integers.

$(x - \frac 12)(x -(-1))$ will be a quadratic with those roots[*], but it might not have integer coefficients. To get integer coefficients we can multiply it by some constant. ... so what $ax^2 + bx + c$ has $\frac 12$ and $-1$ as roots?

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[*] If $x=\frac 12$ then $x-\frac 12= \frac 12 -\frac 12 = 0$ and so $(x-\frac 12)(x-(-1)) =0(x-(-1))= 0$. ANd if $x =-1$ then $(x-(-1)) = 0$ and $(x-\frac 12)(x-(-1)) = (x-\frac 12)\cdot 0 = 0$.

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If a quadratic has roots $r_1$ and $r_2$ then $r_1, r_2$ are solutions to $(x-r_1)(x-r_2)$ and any quadratic that has those to solutions can be factored to $k(x-r_1)(x-r_1)$ where $k$ is some non-zero constant.

So if $ax^2 + bx + c$ has roots $r_1, r_2$ then $ax^2 + bx + c = k(x-r_1)(x-r_2)$.

SO we need to figure out if $k(x-\frac 12)(x - (-1))=ax^2 + bx + c$ and $a,b,c$ are all integers... what are $a,b$ and $c$.

And ... just do it.

$k(x-\frac 12)(x+1) = k(x^2 -\frac 12x + x - \frac 12)= k(x^2 +\frac 12x -\frac 12)=kx^2 + \frac 12kx -\frac 12k$.

So we need $k, \frac 12k, -\frac 12k$ to be integers.

We can do that if $k=2$.

So $2x^2 +x -1$ will have as solutions $\frac 12$ and $-1$.

fleablood
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