Solve the equation by completing the square.?

Question

x2=x square
19.)x2+6x=7
20.)x2+4x=5
21.)x2-8x=9
22.)x2-12x+20=0
23.)x2-4x+2=0
24.)x2+4x-1=0
25.)x2+6x-4=0
26.)x2-2x-5=0
27.)x2+10x-2=0

*Can you guys please give me the answers to these problems?PLEASE

Answer 1

19)x2+6x-7=0
x2+7x-x-7=0
(x+7)(x-1)=0
x=-7 or x=1

20)x2+4x-5=0
a=1, b=4 c=-5
D=4*4-4(1)(-5)
= 16+20
=36
the roots r -4+or-sq root 36
---------------------
2*1
-4+6/2 , -4-6/2
1, -5


21)similarly roots r, x=9 or x=-1

22)x=2 or x= 5

23) 2+sq root2, 2-sq root2

24) -4+sq root17, -4-sq root17
----------------- ----------------
2 2

25) -3+sq root13 , -3-sq root13


26) 1+sq root6 , 1-sqroot6

27) -5+sq root27 , -5-sq root27


oh god! worked a lot for this, hope u give me 10 points by the best answer.

Answer 2

Can't give you the answer, but here is how you do those:

Each equation is of the form ax2 + bx + c = 0. So in x2+6x - 7, a=1, b=6, c= -7.

This equation has 2 solutions x+ and x-

x+ = (-b + sqrt(b2 - 4ac)) / 2a
x- = (-b - sqrt(b2 - 4ac)) / 2a

So just substitute the numbers into the equations and you get 2 solutions per equation.

Good luck.

Answer 3

here's how you do it

23. x2-4x+2=0

subtract 2 from both sides to move it to the other side
x2-4x = -2

take half of the x's coefficient and square it, add to both sides
4/2=2
2^2=4, so

x2-4x+4=-2+4
x2-4x+4=2

factor....
(x-2)^2=2

take the square root of both sides
x-2= +/- sqrt 2

add 2 to both sides to solve for x
x= 2+sqrt 2 or 2-sqrt 2

do the same for the rest of the problems and you should be fine