If x^2 + bx + 5 = (x + c)^2 for all numbers x, where b and c are posit

Question

If  x^2 + bx + 5 = (x + c)^2  for all numbers x, where b and c are positive constants, what is the value of b ?

A.  \sqrt{5}

B.  \sqrt{10}

C.  2\sqrt{5}

D.  2\sqrt{10}

E. 10

PS29980.02

nick1816 · Accepted Answer

x^2+bx +5 = x^2+2cx+c^2

Comparing both sides

c^2= 5

c= \sqrt{5} , as c>0

Also, b= 2c

b = 2\sqrt{5}

AnirudhaS · Answer

Sure.

The problem says for any value of x this holds. So this should hold for x=0, lets put that in.
 0^2 + b.0 + 5 = (0 + c)^2 
or,  5=c^2 
or,  c=\sqrt{5}

Now, putting the value of c back in equation
 x^2 + bx + 5 = (x + c)^2 
or,  x^2 + bx + 5 = (x + \sqrt{5})^2 
or,  x^2 + bx + 5 = x^2 + 2\sqrt{5}x + 5 
or,  bx=2\sqrt{5}x 
or,  b=2\sqrt{5}

lacktutor · Answer

x^2 + bx + 5 = x^{2} + 2xc +c^{2}

-->  c^{2} = 5 
Since  c > 0 , -->  c =√5

b = 2c = 2√5

Answer (C)

kokus · Answer

Can someone share the algebra behind deriving that c^2 = 5? What happens to bx and 2cx? Thanks.

kokus · Answer

Thank you, that makes sense. I thought there is a way to solve without plugging in any value, that's way I was confused. The key apparently is to appreciate that either side must be zero and from there find c.

Rikhraj1 · Answer

AnirudhaS  i didn't understand why u took x=0 ?

Rikhraj1 · Answer

Bunuel  pls explain me this one

Bunuel · Answer

I'd essentially, solve the way nick1816 or AnirudhaS demonstrated above.

For example:

Since  x^2 + bx + 5 = (x + c)^2  for all numbers x, it must be true for x = 0 too, which would result in  c^2 = 5 , which gives  c=\sqrt{5} .

Similarly, since  x^2 + bx + 5 = (x + c)^2  for all numbers x, it must also be true for  x=-\sqrt{5} , which would result in  5 - b*\sqrt{5} + 5 = 0 . This gives  \sqrt{5}b=10 , multiplying by  \sqrt{5}  gives  5b=10\sqrt{5} , and finally  b=2\sqrt{5} .

Answer: C

Rikhraj1 · Answer

Bunuel  i didn't understand how x=0, why not any other no.?

Bunuel · Answer

Check the highlighted part:

If  x^2 + bx + 5 = (x + c)^2     for all    numbers x  , where b and c are positive constants, what is the value of b ?
 
We are given that  x^2 + bx + 5 = (x + c)^2  is true for ALL x, so it must be true for any x you wish, so for  x = 0  too.

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If x^2 + bx + 5 = (x + c)^2 for all numbers x, where b and c are posit

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If x^2 + bx + 5 = (x + c)^2 for all numbers x, where b and c are posit

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