Given that (p - 3 )^2 - 5=0, what is p?

Question

A.  \sqrt{5}+3  or  -\sqrt{5}+3 .

B.  \sqrt{5}+3

C. 2

D.  -\sqrt{5}

E.  -\sqrt{5}+3

TTeja · Answer

p^2 - 6p + 9 - 5 = 0
p^2-6p + 4 = 0

Roots for above equation are irrational.

(OR)

p-3 = '\sqrt{5}'
p = 3 + \sqrt{'5'}

There should be two roots for this equation as well

My apologies I am unable to represent square root function!!

Bunuel · Answer

(p - 3 )^2 - 5=0 ;

(p - 3 )^2 = 5 ;

(p - 3 ) = \sqrt{5}  OR  (p - 3 ) = -\sqrt{5} ;

p = \sqrt{5}+3  OR  p = -\sqrt{5}+3 .

Bunuel · Answer

Writing Mathematical Formulas on the Forum: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 Hope it helps.

akshdeep23 · Answer

I would like to know why I can't solve it like this:
(p-3)^2 -5 =0
(p-3)^2 = 5
p^2-6p+9 =5
p^2-6p = -4
p(p-6)=-4
p=-4
p-6=-4
p=2

Why is solving this way incorrect?

generis · Answer

That's an easy mistake to make. You just forgot that the zero product rule (if ab = 0, then a = 0 or b = 0 or both = 0) ONLY works when the right hand side is zero. You can't set factors in an equation equal to anything but 0 in the way you're solving. You don't have 0 on the right hand side. Plug your answer p = 2 into the original equation; it doesn't work precisely because the zero product rule only applies when the product is zero. Hope that helps! :-)

ydmuley · Answer

(p - 3 )^2 - 5=0

(p - 3)^2 = 5

(p - 3) = +/- \sqrt{5}

p = \sqrt{5} + 3

p = - \sqrt{5} + 3

tapabrata · Answer

(p-3)^2-5=0 
 (p-3)^2=5

(p-3)=+\sqrt{5}  or  (p-3)=-\sqrt{5}

so  p=3+\sqrt{5}  or  p=3-\sqrt{5}

pushpitkc · Answer

For  (p-3)^2-5=0 ,  (p-3)^2  must be equal to 5.

So, p must a number which reduces 3 from the expression  (p-3)^2 
and make its value 5.

There are 2 possible values whose square is 5. 
It is + \sqrt{5}  or - \sqrt{5}

Hence possible values of p must be + \sqrt{5} +3 (or) - \sqrt{5} +3

ccooley · Answer

I've bolded the section where you went wrong. You assumed that because p(p-6) = -4, then p = -4 or p-6 = -4. However, that isn't true. If p(p-6) = -4, you can't conclude that p or p-6 must be -4. (Plug in p = -4 to see why: (-4)(-4-6) actually doesn't equal -4.) The reason you're allowed to do that when the right side of the equation equals 0, is because of the 'zero product rule' that genxer123 mentioned above.

In order to solve it in that way, you'd have to do something kind of convoluted.  Don't solve the problem like this!!  It's very tough to figure this out if you don't already know what the answer is. So, it's just for show:

(p-3)^2 -5 =0

(p-3)^2 = 5

p^2-6p+9 =5

p^2-6p + 4 = 0

(p - 3 - \sqrt{5})(p - 3 + \sqrt{5}) = 0

p=3+\sqrt{5}  or  p=3-\sqrt{5}

bumpbot · Answer

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Given that (p - 3 )^2 - 5=0, what is p?

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Given that (p - 3 )^2 - 5=0, what is p?

Prep Toolkit

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