How do you find the c that makes the trinomial $x^{2} - 7 x + c$ $x^2-7x+c$ a perfect square?

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How do you find the c that makes the trinomial $x^{2} - 7 x + c$ $x^2-7x+c$ a perfect square?

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Answer 1 · 2017-08-19T13:25:33

$c = \frac{49}{4}$ $c=49/4$

Explanation:

${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a - b)^2 = a^2 - 2ab + b^2$
In the above see the relation between middle term and the last term: What do you need to do to $- 2 a b$ $-2ab$ to become $b^{2}$ $b^2$ ?
divide by $2 a$ $2a$ then square the results:
$\frac{- 2 a b}{2 a} = - b$ $(-2ab)/(2a) = -b$ => square: $b^{2}$ $b^2$
Now in this case:
$x^{2} - 7 x + c$ $x^2 - 7x + c$
$c = {(\frac{7}{2})}^{2} = \frac{49}{4}$ $c = (7/2)^2 = 49/4$

Answer 2 · 2017-08-19T13:27:18

$c = \frac{49}{4} = 12 \frac{1}{4}$ $c=49/4=12 1/4$

Explanation:

If a trinomial of the form $x^{2} + b x + c$ $x^2+bx+c$ is a perfect square, then its determinant $b^{2} - 4 c$ $b^2-4c$ must be equal to zero.

In $x^{2} - 7 x + c$ $x^2-7x+c$ , determinant ${(- 7)}^{2} - 4 \times c = 0$ $(-7)^2-4xxc=0$ means

$49 - 4 c = 0$ $49-4c=0$

i.e. $4 c = 49$ $4c=49$

or $c = \frac{49}{4} = 12 \frac{1}{4}$ $c=49/4=12 1/4$ ,

and then $x^{2} - 7 x + \frac{49}{4} = {(x - \frac{7}{2})}^{2}$ $x^2-7x+49/4=(x-7/2)^2$

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How do you find the c that makes the trinomial $x^{2} - 7 x + c$ $x^2-7x+c$ a perfect square?

2 Answers

Explanation:

Explanation:

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