SOLUTION: I am trying to solve the quadratic equation using the method from India. I need assistance with each step. The problem is 2x^2-3x-5=0 There are 6 steps and I have gotten

Question 577908: I am trying to solve the quadratic equation using the method from India.
I need assistance with each step.
The problem is 2x^2-3x-5=0
There are 6 steps and I have gotten to step 3
Step 1 2x^-3x=5
Step 2 8x^2-12x=20
Step 3 -3 is the original x term, therefore:
8x^2-12x+9=29
Step 4, now I am to take the square root of each side. This is where I am confused, need help here through step 6.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The problem is 2x^2-3x-5=0
There are 6 steps and I have gotten to step 3
Step 1 2x^2-3x=5
Question: Why do you multiply by 4?
Step 2 8x^2-12x=20
Step 3 -3 is the original x term, therefore:
8x^2-12x+9=29
-----
I don't think your 3rd step is correct.
According to your step 4 you need to complete the
square in step 3. That would be as follows:
8x^2-12x = 20
x^2 - (3/2)x = (5/2)
Competing the square you would get:
x^2 - (3/2)x + (3/4)^2 = (5/2) + (3/4)^2
Now you are ready to factor and then take the square root of both sides.
-----------------------
Step 4, now I am to take the square root of each side.
(x-(3/4))^2 = (49/16)
(x-(3/4)) = +-7/4
x = (7/4)+(3/4) or x = -(7/4)+(3/4)
x = 5/2 or x = -1
======================
Cheers,
Stan H.
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