How do you find the value of c that makes $x^{2} + 5 x + c$ $x^2+5x+c$ into a perfect square?

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Answer 1 · 2017-02-20T19:48:57

$c = \frac{25}{4}$ $c = 25/4$

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Method 1

Note that ${(x + k)}^{2} = x^{2} + 2 k + k^{2}$ $(x+k)^2 = x^2+2k+k^2$

So if $2 k = 5$ $2k = 5$ then $k = \frac{5}{2}$ $k = 5/2$ and:

${(x + \frac{5}{2})}^{2} = x^{2} + 5 x + {(\frac{5}{2})}^{2} = x^{2} + 5 x + \frac{25}{4}$ $(x+5/2)^2 = x^2+5x + (5/2)^2 = x^2+5x+25/4$

So $c = \frac{25}{4}$ $c = 25/4$

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Method 2

$x^{2} + 5 x + c$ $x^2+5x+c$

is in the form:

$a x^{2} + b x + c$ $ax^2+bx+c$

with $a = 1$ $a=1$ and $b = 5$ $b=5$ .

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = 25-4c$

So if $Delta = 0$ (indicating a repeated zero) then $25-4c = 0$ and hence $c = 25/4$ .

How do you find the value of c that makes x^2+5x+cx2+5x+cx^2+5x+c into a perfect square?