How do you solve 3( x - 2 ) ^ { 2 } - 2 0 = 5 5?

1 Answer
Oct 12, 2016

x_1 = 7
x_2 = -3

Explanation:

This equation can be solved in a number of ways, but I think this way is the simplest and leads to the fewest errors.

1.Evaluate the brackets:

3(x-2)^2-20=55

In order to evaluate (x-2)^2, we just need to remember that any number squared can just be written as being multiplied by itself.

3(x-2)(x-2)-20=55

We can evaluate (x-2)(x-2) using the FOIL method

3(x^2-4x+4)-20=55

2.Expand the brackets:

3(x^2-4x+4)-20=55
3x^2-12x+12-20=55

3. Set the equation equal to 0 and collect like terms:

3x^2-12x+12-20=55
3x^2-12x+12-20-55=0
3x^2-12x-63=0

4.Solve for the unknown:

3x^2-12x-63=0

The easiest way to solve this is to have a coefficient of 1 for x^2, as this will make factorising the equation easier:

x^2-4x-21=0

At this stage, we have a simplified equation equal to 0. In order to find the roots of x, we need to find two factors which multiply to give c (in this case -21) and add to give b (in this case -4):

x^2-7x+3x-21=0

Then we factorise each 'pair' of terms in the equation:

x(x-7)+3(x-7)=0

Since we are multiplying (x-7) by x and by +3, we can put these two multipliers in one bracket to make solving the equation easier:

(x-7)(x+3)=0

5.Set each bracket equal to 0 to get values for x:

Since we know that if a*b=0, then a or b must be equal to 0, we can apply that rule here. But since we don't know which factor equals 0, we set both equal to 0 and solve for both.

x-7=0
x=7

x+3=0
x=-3

x_1 = 7
x_2 = -3