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I have the equation- $(-3p)^2 + 4(4p+1)$

how do I make it have equal roots? cause I don't think it's possible but must be, can someone please help

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    @Siobhan: Suppose that instead of what you wrote down, the actual problem is to solve $(-3p)^2-4(4p+1)=0$. This equation simplifies to $9p^2-16p -4=0$, which can be rewritten as $(9p+2)(p-2)=0$. Then the roots are $-2/9$ and $2$. Your post put us in the middle of a problem. Can you restate the problem exactly as it was asked?2012-02-28

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If I understand correctly all you need is to factorize this polynomial. You get by solving the 2nd order equation $9p^2+16p+4=0$: $9p^2+16p+4=-\frac{1}{9}(-9p+2\sqrt{7}-8)(9p+2\sqrt{7}+8)$

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    You're not a lost cause! You just need to be a bit more careful, and also learn the correct terminology.2012-02-28
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A root of an polynomial is a value for the variable that makes the polynomial equal to zero. So you can't "make it have equal roots", it has the roots that it has and they won't change.

The roots happen to be $\dfrac {2} {9}\left(-4+\sqrt {7}\right)$ and $\dfrac {2} {9}\left( -4-\sqrt {7}\right)$. These are not equal. There is a different kind of problem where some of the numbers in your polynomial are unknown and you are expected to find a value for those that will give multiple roots that are equal, but that would be a different problem.