Help!!!!!!! How do you do completing the square with these problems?

Question

1. x^2-13x+36+0

2. x^2+3x=0

3. x^2=-11

Answer 1

Use the quadratic equation formula -b +/- the square root b^2 -4(a)(c) divide by 2(a)

13 +/- sq rt 13^2-4(1)(36)
-----------------------------------
2(1)

And solve 13+/- sq root 25 so.. 13+ 5 = 18 divided by 2 = 9
13- 5 = 8 divided by 2 = 4

And just plug in those answers to see if it equals 0 to double check your answers. And then use the equation for # 2. Hope this helps

Answer 2

1)

x^2 - 13x + 36 = 0

x^2 - 2(13/2) x + (13/2)^2 - (13/2)^2 + 36 = 0

(x - 13/2)^2 + 36- 169/4 =0

(x- 13/2)^2 -25/4 = 0

(x-13/2)^2 = (5/2)^2

x = 13/2 +/- 5/2

= (13+5)/2 or (13 - 5)/2

= 9, 4

2)

x^2 + 3x = 0

x^2 + 2(3/2)x + (3/2)^2 - (3/2)^2 = 0

(x + 3/2)^2 = (3/2)^2

x + 3/2 = +/- 3/2

x = -3/2 +/- 3/2

= - 3 or 0

3)

x^2 = -11


x = +/- sqrt(-11)

= +/- sqrt(11)i

Answer 3

1. half of 13 is 6.5, 6.5 squared is 42.25. add it into the part you want to complete the square with, and subtract from the part you aren't, so (x^2-13x+42.25) -6.25. solve the rest.

2. half of 3 is 1.5, squared is 2.25, so your entire equation is (x^2+3x+2.25) -2.25. solve the rest.

3. can't do last one. a squared number (x^2) can't be negative (-11)

Answer 4

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