Can somebody help me with algebra I math homework on "completing the square?"?

Question

Use the most convieninet method to solve the equation.
2x^2 + x - 3

2x^2 + 8x + 3 = 10

3^2 + 4x - 2 = 0

5x^2 - x - 2 = 0

6x^2 + 11x = -3

x^2 + 5 = 7x

x^2 indicates squared or to the whatever power.

HElp!

the answer format is supposed to be like x = 3 + or - the square root of 6. something like that.

Answer 1

2x^2 + x - 3 = 0
2x^2 + x = 3
x^2 + (1/2)x = (3/2)
x^2 + (1/2)x + (1/16) = (25/16)
(x + (1/4))^2 = (25/16)
x + (1/4) = ±(5/4)
x = (-1/4) ± (5/4)
x = (-6/4) or (4/4)

ANS : (-3/2) or 1

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3x^2 + 4x - 2 = 0
3x^2 + 4x = 2
x^2 + (4/3)x = (2/3)
x^2 + (4/3)x + (4/9) = (10/9)
(x + (2/3))^2 = (10/9)
x + (2/3) = ±(1/3)sqrt(10)
x = (-2/3) ± (1/3)sqrt(10)

ANS : (1/3)(-2 ± sqrt(10))

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5x^2 - x - 2 = 0
5x^2 - x = 2
x^2 - (1/5)x = (2/5)
x^2 - (1/5)x + (1/100) = (41/100)
(x - (1/10))^2 = (41/100)
x - (1/10) = ±(1/10)sqrt(41)
x = (1/10) ± (1/10)sqrt(41)

ANS : (1/10)(1 ± sqrt(41))

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6x^2 + 11x = -3
x^2 + (11/6)x = (-1/2)
x^2 + (11/6)x + (121/144) = (49/144)
(x + (11/12))^2 = (49/144)
x + (11/12) = ±(7/12)
x = (-11/12) ± (7/12)
x = (-18/12) or (-4/12)

ANS : (-3/2) or (-1/3)

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x^2 + 5 = 7x
x^2 - 7x = -5
x^2 - 7x + (49/4) = (29/4)
(x - (7/2))^2 = (29/4)
x - (7/2) = ±(1/2)sqrt(29)
x = (7/2) ± (1/2)sqrt(29)

ANS : (1/2)(7 ± sqrt(29))

Answer 2

well...it's pretty ez if u think bout it. u take the middle #...( the 1 w/ the x) and don't worry bout the sign. divide that # by 2 and then square it. then add that # to both sides ( btw..u should have the # w/out the x on the other side of the = sign. ) then u take x and the sign of the middle # and the # from up there (b4 u square the eq) and take square root of the # on the other side of the = sign. lol...let me show u...here's an example.

x^2-6x=40...(set it up like this)
then take 6/2. which is 3. then square it. which is 9.
add 9 to both sides.
x^2-6x+9=40+9
now take x and the sign in the middle (-) and the 3 b4 the squaring(3)
so x-3
now square root the other side of the =.
+- 7 (it's plus or minus cuz it's a square root)
so now u have
x-3=+-7
u can do that now cant cha

Answer 3

the first component you go with to do in each and every of the questions is positioned the same words on the same area of the equivalent signal. So for the first one: 3x - 5 = 6x -14 positioned the x's on the same area and the non x's on the same area i.e. 3x - 6x = -14 + 5 (each and every time you bypass a variable from one area of the = signal to the different it differences even if it is constructive or adverse) So for this one you should get : -3x = -9 to locate x you divide -9 through -3. X = 3 For the second one one: 12n - 5 = 4n + 35, again positioned the n's on the same area and the non n's on the different area 12n - 4n = 35 + 5. this supplies 8n = 40 To get n divide 40 through 8 giving : N = 5 For the 0.33 you first ought to multiply out the bracket: 2( t + 7) = 5t -7 turns into: 2t + 14 = 5t - 7. again, positioned the t's on the same area: 2t - 5t = -7 - 14 Giving: -3t = -21 to locate t divide -21 through -3 hence : t = 7

Answer 4

2x^2+x-3=0
=>2x^2+3x-2x-3=0
=>x(2x+3)-1(2x+3)=0
=>(2x+3)(x-1)=0
hence eithr 2x+3=0 or x-1=0
Therefore x= -3/2 or 1
2x^2+8x+3=10
2x^2+8x=10-3
2x^2+8x=7
=>x^2+4x=7/2 [dividing all the trms by 2]
=>(x)^2+2.x.2+(2)^2=7/2+4
=>(x+2)^2=15/2
=>x+2=+-sqrt(15/2)
=>x=-2+-sqr(15/2)

3^2+4x-2=0
9+4x-2=0
4x=-7
=>x= -7/4
5x^-x-2=0
=>5x^2-x=2
=>x^2-x/5=2/5 [dividing all terms by 5]
=(x)^2-2,x.1/10+(1/10)^2=2/5+(1/10)^2
=>(x-1/10)^2=2/5 +1/100=41/100
=>x-1/10=+-(sqrt)41/100=(+-sqrt41)/10
=>x=1/10 +-(sqrt41)/10
=(1+-sqrt41)/10

6x^+11x=-3
=>x^2+11x/6=-1/2 [dividing all the terms by 6]
=>(x)^2+2.x.11/12+(11/12)^2= -1/2+(11/12)^2
=>(x+11/12)^2= -1/2+121/144=49/144
=>x+11/12=sqrt 49/144=+-7/12
=>x=-11/12+-7/12
=-4/12 or -1/3 and -18/12 or -1 1/2
x^2+5=7x
=>x^2-7x=-5
=>(x)^2-2.x.(7/2)+(7/2)^2= -5+(7/2)^2
=>(x-7/2)^2=-5+49/4=29/4
=>x-3 1/2=sqrt 29/4=+- (sqrt29)/2
x=3 1/2+- (sqrt29)/2

Answer 5

1) (x-1)(2x+3)

2) 2x^2+8x-7=0
(-8(+-)sqrt(64+56))/4=x
(-8(+-)sqrt(120))/4=x
(-8(+-)2rt30)/4=x

3)(-4(+-)sqrt(16+24))/6
(-4(+-)2rt10)/6

plug in the quadratic formula...: x=(-b(+-)sqrt(b^2-4ac))/2a
the equations are formatted like: ax^2+bx+c=0

Answer 6

look in ur math book it should explain how to do what your doing 1 or 2 pages before your homework