Please answer all questions and you will get brainliest:

For what real values of $c$ is $x^2 - 8x + c$ the square of a binomial?

For what real values of $s$ is $x^2 + sx + 144$ the square of a binomial?

For what real values of $c$ is $4x^2 + 14x + c$ the square of a binomial?

Respuesta :

Question 1

We want to find real values of c, for which

[tex] {x}^{2} - 8x + c[/tex]

is the square of a binomial.

If the above quadratic equation is the square of a binomial, then its discriminant is zero.

This implies that:

[tex]D = {b}^{2} - 4ac = 0[/tex]

where a=1, and b=-8

We substitute the values into the formula:

[tex] {( - 8)}^{2} - 4(1)(c) = 0[/tex]

[tex]64- 4c= 0 \\ 4c = 64 \\ c = 16[/tex]

Question 2

We want to find the value of s for which

[tex] {x}^{2} + sx + 144[/tex]

is the square of a binomial.

In this case too the discriminant must be zero.

D=b² -4ac=0

We put a=1,b=s, and c=144 to obtain:

[tex] {s}^{2} - 4(1)(144) = 0[/tex]

[tex] {s}^{2} - 576 = 0[/tex]

[tex] {s}^{2} = 576[/tex]

[tex]s = \pm \sqrt{576} [/tex]

[tex]s = \pm24[/tex]

Question 3

We want to find the value of c, for which

[tex]4 {x}^{2} + 14x + c = 0[/tex]

Again we have:

[tex] {b}^{2} - 4ac = 0[/tex]

We put a=4 and b=14

[tex] {14}^{2} - 4(4)(c) = 0[/tex]

[tex]196 - 16c = 0[/tex]

[tex]16c = 196[/tex]

[tex]c = \frac{49}{4} [/tex]

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