Solve each equation by completing the square.
-36x=8x^2+28
-8x^2-3x=-5
5x^2-x-6=0
6x^2-24x-30=0
7x^2+8x-12=0
-10x^2 -9x+1=0
5x^2-x=4
-11x^2-8x=-7x-10
-3 1/3x^2+ 10x -27 1/2= -231/3
#7) 5x^2-x=4 is x=1
36x = 8x^2 +28
8 x^2 -36x +28 = 0
x^2- (9/2)x + 7/2 = 0
(x - 9/4)^2 +7/2 - 81/16 = 0
(x - 9/4)^2 = (81-56)/16 = 25/16
x - 9/4 = + or - 5/4
x = 1 or 7/2
I'll be glad to check your work on the other ones
To solve each equation by completing the square, follow the steps below:
Step 1: Rearrange the equation so that the variable terms are on one side and the constant term is on the other side.
Step 2: Divide the equation by the coefficient of the squared term to make the coefficient 1.
Step 3: Move the constant term to the other side of the equation.
Step 4: Take half of the coefficient of the linear term, square it, and add it to both sides of the equation.
Step 5: Factor the trinomial on the left side of the equation, if possible.
Step 6: Solve for the variable using the square root property.
Now let's solve each equation:
1) -36x = 8x^2 + 28
Rearrange the equation: 8x^2 + 36x + 28 = 0
Divide by the coefficient of the squared term: x^2 + 4.5x + 3.5 = 0
Move the constant term to the other side: x^2 + 4.5x = -3.5
Take half of the coefficient of the linear term, square it, and add it to both sides: x^2 + 4.5x + (4.5/2)^2 = -3.5 + (4.5/2)^2
x^2 + 4.5x + 2.25 = -3.5 + 2.25
Factor the trinomial: (x + 2.25)^2 = -1.25
Take the square root of both sides: x + 2.25 = ±√(-1.25)
Since the square root of a negative number is not a real number, there is no real solution for this equation.
2) -8x^2 - 3x = -5
Rearrange the equation: 8x^2 + 3x - 5 = 0
Divide by the coefficient of the squared term: x^2 + (3/8)x - (5/8) = 0
Move the constant term to the other side: x^2 + (3/8)x = (5/8)
Take half of the coefficient of the linear term, square it, and add it to both sides: x^2 + (3/8)x + (3/16)^2 = (5/8) + (3/16)^2
x^2 + (3/8)x + 9/256 = 40/256 + 9/256
Factor the trinomial: (x + 3/16)^2 = 49/256
Take the square root of both sides: x + 3/16 = ±√(49/256) = ±(7/16)
Solve for x: x = -3/16 ± (7/16)
Therefore, the solutions are x = -10/16 and x = 4/16, which simplify to x = -5/8 and x = 1/4.
3) 5x^2 - x - 6 = 0
Divide by the coefficient of the squared term: x^2 - (1/5)x - (6/5) = 0
Move the constant term to the other side: x^2 - (1/5)x = (6/5)
Take half of the coefficient of the linear term, square it, and add it to both sides: x^2 - (1/5)x + (1/10)^2 = (6/5) + (1/10)^2
x^2 - (1/5)x + 1/100 = 30/100 + 1/100
Factor the trinomial: (x - 1/10)^2 = 31/100
Take the square root of both sides: x - 1/10 = ±√(31/100) = ±(√31/10)
Solve for x: x = 1/10 ± (√31/10)
The solutions are x = (1 ± √31)/10.
Continue to solve the remaining equations using the steps outlined above.