Solve each of the following quadratic equations by completing the square.?

Question

1#
x^2-6x-3=0

2#
2x^2+10x+11=0

Answer 1

x^2-6x+9=3+9
(x-3)^2=+/-rt12
x-3=+/-2rt3
x=3+/-2rt3

2.2(x^2+5x)=-11
2(x^2+5x+25/4-25/4)=-11
2(x+5/2)^2-25/2=-22/2
2(x+5/2)^2=3/2
(x+5/2)^2=3/4
x+5/2=+/-(rt3)/2
x=-5/2+/-(rt3)/2
=1/2[-5+/-rt3]

Answer 2

polishing off the sq. is powerful in circumstances like this which have not were given a speedy factoring answer. good that is the first one. x^2 - 6x - 3 = 0. step one is to isolate the x words and the constants. So, good the following, upload 3 to both factors. x^2 - 6x = 3. Now, to end the sq., we upload to both factors the quantity which will enable us to ingredient the left ingredient in to a spread of (ax + b)^2. This quantity is (6/2)^2. Why? shall we favor to be waiting to characteristic b to itself to get 6. in diverse words, 2b = 6. So, our equation turns into: x^2 - 6x + (6/2)^2 = 3 + (6/2)^2. Simplifying resources us: x^2 - 6x + 9 = 12 (x - 3)^2 = 12 x - 3 = +/- sqrt(12) x = 3 +/- 2(sqrt(3)). the second one contains an extra effective step by creating use of first time period being better by 2. Set it up as follows. 2x^2 + 10x + 11 = 0 2x^2 + 10x = (-11) x^2 + 5x = (-5.5) or (-11/2). Then, keep on with the technique concentrated above. x^2 + 5x + (5/2)^2 = (-11/2) + (5/2)^2 x^2 + 5x + 25/4 = (-22/4) + (25/4) (x + (5/2))^2 = 3/4 x + 5/2 = +/- (sqrt(3))/2 x = (-5/2) +/- (sqrt(3))/2 = (-5 +/- sqrt(3))/2.

Answer 3

Completing the square is helpful in situations like this that don't have a quick factoring solution. Here's the first one.

x^2 - 6x - 3 = 0.

The first step is to isolate the x terms and the constants. So, here, add 3 to both sides.

x^2 - 6x = 3.

Now, to complete the square, we add to both sides the number that will allow us to factor the left side in to a form of (ax + b)^2. This number is (6/2)^2. Why? We must be able to add b to itself to get 6. In other words, 2b = 6.

So, our equation becomes:

x^2 - 6x + (6/2)^2 = 3 + (6/2)^2.

Simplifying gives us:

x^2 - 6x + 9 = 12
(x - 3)^2 = 12
x - 3 = +/- sqrt(12)
x = 3 +/- 2(sqrt(3)).

The second one involves one extra step because of the first term being multiplied by two. Set it up as follows.

2x^2 + 10x + 11 = 0
2x^2 + 10x = (-11)
x^2 + 5x = (-5.5) or (-11/2).

Then, follow the procedure detailed above.

x^2 + 5x + (5/2)^2 = (-11/2) + (5/2)^2
x^2 + 5x + 25/4 = (-22/4) + (25/4)
(x + (5/2))^2 = 3/4
x + 5/2 = +/- (sqrt(3))/2
x = (-5/2) +/- (sqrt(3))/2 = (-5 +/- sqrt(3))/2.

Answer 4

To complete the square of quadratic function:y= ax^2+bx+c
Let y= d(x+e)^2+f be the function in completed square form

d=a
e=b/2a
f=c-e^2

#1.y=(x-3)^(2) -3 -9 =(x-3)^(2) - 12=0
Now solve for x:
y=(x-3)^(2)=12
x-3=+ or - square root of 12= +/- 2sqrt(3)
x= 3 +/- 2sqrt (3)

#2. y=2(x+5)^(2)+ 11-2*(5/2)^(2)=2(x+5/2)^(2) - 3/2=0

2(x+5/2)^(2) =3/2
(x+5/2)^(2) =3/4
x+5/2= +/-sqrt(3/4)= +/- (1/2)*sqrt(3)
x= -5/2 +/- (1/2)*sqrt(3)= (1/2)*(-5 +/- sqrt 3)

Answer 5

#1
x= [6+/- rt(36+12)]2=3+/- rt48 /2=3+/- 2rt3

#2
x= [-10 +/- rt(100-88)]/4 =[-5 +/- rt3]/2

Answer 6

1.
x^2-6x-3
=x^2-6x+9-9-3
=[x-3]^2-12
=[x-3]^2-[2sqrt3]^2
=[x-3+2sqrt3][x-3-2sqrt3]
=0
x=3-2sqrt3
x=3+2sqrt3
2.
2x^2+10x+11
=2[x^2+5x+5.5]
=2[x^2+5x+6.25-6.25+5.5]
=2[(x+2.5)^2-.75]
=2[(x+2.5)^2-(sqrt.75)^2]
=2[(x+2.5+sqrt.75)(x+2.5-sqrt.75)]
=0
x=-2.5-sqrt.75
x=-2.5+sqrt.75