How do you find the value of c that makes $x^{2} + 26 x + c$ $x^2+26x+c$ into a perfect square?

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Answer 1 · 2018-06-09T17:07:57

$c = 169$ $c=169$

square is ${(x + 13)}^{2}$ $(x+13)^2$

In the form:

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

to complete the square $a = 1$ $a=1$ is required.

$c = {(\frac{b}{2})}^{2}$ $c = (b/2)^2$

the square term is $\frac{b}{2}$ $b/2$

$x^{2} + 26 x + c$ $x^2+26x+c$

$a = 1$ $a=1$ so we are okay to complete the square.

$b = 26$ $b=26$

$c = {(\frac{26}{2})}^{2} = 13^{2} = 169$ $c=(26/2)^2=13^2=169$

finally to complete the square:

$x^{2} + 26 x + 169 = {(x + \frac{b}{2})}^{2} = {(x + 13)}^{2}$ $x^2+26x+169=(x+b/2)^2 = (x+13)^2$

How do you find the value of c that makes x2+26x+cx^2+26x+c into a perfect square?