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Solve Each Equation By Completing the Square. Please show me or help me
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# Solve Each Equation By Completing the Square. Please show me or help me

[From: ] [author: ] [Date: 11-05-16] [Hit: ]
(x-6)(x-8)x=6,3. (x+10)(x+2)x=-2,5.p^2-8p+15=0 (p-5)(p-3)=0p=3,6 (a-4)(a++)=0a=4,......
1) k^2 + 8k + 12 = 0

2) a^2 - 2a - 48 = 0

3) x^2 + 12x + 20 =0

4) m^2 - 12m + 26 = 0

5) p^2 - 8p +21 = 6

6) a^2 - 2a - 8 = 0

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Half the middle term in this case 8k/2 is 4k
then 4 goes into the brackets, 4^2 is 16 and we need it to equal 12 so outside the brackets put -4
(k+4)^2-4=0
From this then rearrange the formula to equal k
(k+4)^2=4
k+4=+/-2
k= -2 and k=-6

EDIT to check you can substitute k= -2 back into the equation
(-2)^2+8(-2)+12 = 0

2) (a-1)^2-49=0
(a-1)^2=49
a-1=+/-7
a=8 and -6
8^2-2(8)-18=0

3) (x+6)^2-4=0
(x+6)^2=4
x+6=+/-2
x=-4 and -8

From this you should be able to get the general gist of it if you don't just message me.

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1. k^2 + 8k +12= 0 what two numbers multipy to be 12 but add to be 8 (6,2)

(k+6)(k+2)=0
k=-6
k=-2

Check answer -2^2 + 8(-2)+ 12 =0 ==> 4-16+12=0 and -6^2=8(-6)+12 = 36-48+12= 0 cks

2. (x-6)(x-8) x=6,8
3. (x+10)(x+2) x=-2,-10
4 did you write 4 wrong
5.p^2-8p+15=0
(p-5)(p-3)=0 p=3,5 ck 3^2-8(3)+15 = 9-24+15 = 0 ck 25-40=15 = 0 ck
6 (a-4)(a++)=0 a=4,-2 ck 16-8-8=0 ck and 4+4-8 =0 ck

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1. k = -2,-6
2. a = 8, -6
3. x = -2, -10
4. m = sqrt.(10) + 6, - sqrt.(10) + 6
5. p = 5, 3
6. a = 4, - 2
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