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4 years ago

15

Write x squared + 2x-8 in the form (x + m) squared + n where m and n are integers.

1 answer:

skelet666 [1.2K]4 years ago

7 0

Let F(x) = x^2 + 2x -8

Adding 1 and - 1 at equation

F(x) = x^2 + 2x + 1 - 1 -8

Rewrinting the first 1 = 1^2

F(x) = x^2 + 2x + 1^2 - 1 - 8

Now look:

x^2 + 2x + 1^2 = ( x + 1)^2

Then,

F(x) = (x + 1)^2 -1 -8

F(x) = (x + 1)^2 -9

This is the answer !

If us to want.

(x + 1)^2 -9 = ( x+1)^2 - 3^2

= ( x+ 1 -3).(x +1 +3)

= ( x - 2).(x +4)

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3 years ago

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3 years ago

H(x) = x2 + 1 k(x) = x – 2 (h + k)(2) = (h – k)(3) = Evaluate 3h(2) + 2k(3) =

matrenka [14]

For this case we have the following functions:

$h (x) = x ^ 2 + 1$

$k (x) = x - 2$

We must perform each of the operations shown.

We have then:

For (h + k) (2):

$(h + k) (x) = h (x) + k (x)$

$(h + k) (x) = (x ^ 2 + 1) + (x - 2)$

$(h + k) (x) = x ^ 2 + x - 1$

Evaluating for x = 2 we have:

$(h + k) (2) = 2 ^ 2 + 2 - 1$

$(h + k) (2) = 4 + 2 - 1$

$(h + k) (2) = 5$

For (h - k) (3):

$(h - k) (x) = h (x) -k (x)$

$(h - k) (x) = (x ^ 2 + 1) - (x - 2)$

$(h - k) (x) = x ^ 2 - x + 3$

Evaluating for x = 3 we have:

$(h - k) (3) = 3 ^ 2 - 3 + 3$

$(h - k) (3) = 9 - 3 + 3$

$(h - k) (3) = 9$

For 3h (2) + 2k (3):

$3h (2) + 2k (3) = 3 (2 ^ 2 + 1) + 2 (3 - 2)$

$3h (2) + 2k (3) = 3 (5) + 2 (1)$

$3h (2) + 2k (3) = 15 + 2$

$3h (2) + 2k (3) = 17$

Answer:

$(h + k) (2) = 5$

$(h - k) (3) = 9$

$3h (2) + 2k (3) = 17$

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3 years ago

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