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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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That's not an equation. An equation mus have an equal sign in it. And if you want to think of it as a polynomial in $p$, then you cannot "make it" do anything. It is a specific, fixed, polynomial; the discriminant is $16^2 - 16\times 9 = 16(7)\neq 0$ (unless you are working over a field of characteristic $2$ or $7$, I guess...) –  Arturo Magidin Feb 28 '12 at 20:53
    
it is =0 (I think) but i just need it to go into 2 bracets –  Siobhan Feb 28 '12 at 20:56
    
"is =0" doesn't make sense to me; and I don't know what "go into 2 bracets" means. –  Arturo Magidin Feb 28 '12 at 21:01
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btw it is English! –  Siobhan Feb 28 '12 at 21:12
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@AndréNicolas' guess gets my vote for what was originally being asked of the OP. Also Siobhan, 1 - "getting to go into two brackets" is called factoring. 2 - $$x^2 + 2$$ does NOT factor into $$(x + 1)(x + 2)$$. You should distribute/"FOIL" this out to check... –  The Chaz 2.0 Feb 28 '12 at 21:23
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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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thank that is almost what i need to do but the answer is much simpler (I'm only doing higher maths). Buts thank for the help –  Siobhan Feb 28 '12 at 21:19
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If "the answer is much simpler", then the question is not what you asked. Could it have been $(-3p)^2 + 4(3p+1)$? –  Robert Israel Feb 28 '12 at 21:25
    
Ok can we ust accept that I've written it wrong and leave it be. Thank again for all the help, but I'm a lost cause I'm afraid –  Siobhan Feb 28 '12 at 21:29
    
You're not a lost cause! You just need to be a bit more careful, and also learn the correct terminology. –  Robert Israel Feb 28 '12 at 21:55
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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.

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