Question

The solution by completing the squares 3x^2-32x+12=0

Answer 1

Given equation,

3x² - 32x + 12 = 0

First we have to divide both sides by the coefficient of x², i.e., 3.

x² - (32/3)x + 4 = 0

Now, subtract the constant term (coefficient of x^0), i.e., 4, from both sides.

x² - (32/3)x = -4

Now, add the square of half the coefficient of x to both sides.

Here the coefficient of x is -32/3.

(-32/3 • 1/2)² = (-16/3)² = 256/9

So we have to add 256/9 to both sides.

x² - (32/3)x + 256/9 = - 4 + 256/9

Now we factorise the LHS and simplify the RHS.

(x - 16/3)² = (256 - 36) / 9

=> (x - 16/3)² = 220/9

Now taking the square root of both sides...

x - 16/3 = ±√(220/9)

=> x - 16/3 = ± (2√55) / 3

Now add 16/3 to both sides, and we get,

x = (16 ± 2√55) / 3