What number should be added to both sides of the equation to complete the square x^2+8x=4

2 answers:

5 0

It would be half of the coefficient of x squared added to both sides:
So half of 8 ( the coefficient of 8x) is 4;
The number 4 squared is 16, so you would add that to both sides.

You do this so that you can easily factor the left side: x^2 + 8x + 16 = 4 + 16
(x + 4)(x+4) = 20

UkoKoshka [18]3 years ago

5 0

Answer:

To complete the square you have to add 20.

Step-by-step explanation:

The concept of square of a binomial can be represented as

(a + b)² = a² + b² + 2ab

Then given x² + 8x = 4 ⇒ x² + 8x -4 = 0. From this expression we can easy identify "a" as x. Then b can be obtained if we think that 2ab=8x

⇒2xb=8x ⇒b = 8x/2x ⇒b=4.

As you can see if a=x and b=4, then you have (x + 4)²

⇒(x + 4)²= x² + 16 + 8x.

Finally if the final expression is x² + 16 + 8x, you have to add 20 to the given expression.

x² + 8x - 4 + 20 = x² + 8x + 16.

You might be interested in

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :