(p^2)(x^2)-12x+p+7=0 has the root 3/2..find the value of p.

please help,show step by step the answer that u get..thanks

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- Feb 26th 2010, 12:38 AMmastermin346root of the equation
(p^2)(x^2)-12x+p+7=0 has the root 3/2..find the value of p.

please help,show step by step the answer that u get..thanks - Feb 26th 2010, 02:49 AMHallsofIvy
- Feb 26th 2010, 02:58 AMBacterius
To explain HallsOfIvy's solution, remember that the roots of an equation are the values of $\displaystyle x$ when $\displaystyle y = 0$. Here, you are given an equation in the form $\displaystyle y = f(x)$ with $\displaystyle y = 0$ and $\displaystyle f(x) = (p^2)(x^2)-12x+p+7$ for some $\displaystyle p$. Since you are given that one root is $\displaystyle \frac{3}{2}$, you know that when $\displaystyle y = 0$ (which is our case), $\displaystyle x = \frac{3}{2}$. So you can just substitute $\displaystyle x = \frac{3}{2}$ into $\displaystyle f(x)$ and solve for $\displaystyle p$, to find which value

__s__of $\displaystyle p$ (there may be more than one) satisfy the equation $\displaystyle f(x) = 0$ when $\displaystyle x = \frac{3}{2}$. :)