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Firlakuza [10]
3 years ago
5

Rewrite x^2 + 6x − 1 = 0 in the form (x + a)^2 = k, where a and k are constants. What is k?

Mathematics
1 answer:
Yuri [45]3 years ago
5 0

answer

10

step-by-step explanation

first find the (x+a)^2 part

using x^2 + 6x + c, we need to find the value c that completes the perfect square

to do this, divide 6 by two to find a in (x+a)^2

6/2 = 3 = a

c = a^2

c = 3^2 = 9

plug in values

x^2 + 6x + 9 = (x+3)^2

compare this to  x^2 + 6x − 1, and you can see there is a (9 - -1) = 10 difference, so subtract 10 from both sides

x^2 + 6x + 9 = (x+3)^2

x^2 + 6x + 9 - 10 = (x+3)^2 - 10

x^2 + 6x + - 1 = 0 = (x+3)^2 - 10

0 = (x+3)^2 - 10

(x+3)^2 = 10

k = 10

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JK=2X KL=X+2 and JK=5X-10
elixir [45]

JK + KL = JL

JK = 2x, KL = x + 2, JL = 5x - 10

therefore we heve the equation:

2x + (x + 2) = 5x - 10

3x + 2 = 5x - 10 |-2

3x = 5x - 12 |-5x

-2x = -12 |:(-2)

x = 6

JK = 2x → JK = 2(6) = 12

KL = x + 2 → KL = 6 + 2 = 8

JL = 5x - 10 → JL = 5(6) - 10 = 30 - 10 = 20

Check:

JK + KL = JL

12 + 8 = 20 CORRECT :)

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