Quadratic Equations

[sully1964]

First please understand that this is the first math class I have had in 25 years. The assignment is as follows:
An interesting method fo solving QE came from India. The steps are for the following problem are: x^2 +3x -10 =0
a) Move the constant term to the right side of the equation - X^2 +3X=10
b) Multiply each term by 4 times the coefficient of the x^2 term - 4x^2 +12x=40
c) Square the coefficient of the original x term and add it to both sides of the equation - 4x^2 +12x+9=40+9 then you get 4x^2+12x+9=49
d) Take the square root of both sides - 2x+3=plus or minus 7 Here is the beginning of my problems. please explain in detail how this is the answer
e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.
f) Set the left side of the equation equal to the negative square root of the number on the right side and solve for x.

2x+3=7 2x+3=-7
2x=4 or 2x=-10
x=2 x=-5

I only have 4 problems to do but cant get past step e) Can someone help?

 

[tkhunny]

It is ALWAYS instructive to use arbitrary coefficients.

ax^2 + bx + c = 0

ax^2 + bx = -c

4(a^2)x^2 + 4abx + b^2 = -4ac + b^2

A step they might have mentoned, but didn't

4(a^2)x^2 + 4abx + b^2 = (2ax + b)^2 = -4ac + b^2 = b^2 - 4ac

Now back to step d.

(2ax + b)^2 = -4ac + b^2 ==> 2ax + b = sqrt(b^2 - 4ac) OR 2ax + b = -sqrt(b^2 - 4ac)

It's pretty much regular algebra from here. No more Indian secrets.

Just another way to "Complete the Square". Note: It is unlikely that our Indian friends saw this as a general way to solve ALL Quadratic Equations. Ah, what great insights a few centuries will provide.