This equation can be solved in a number of ways, but I think this way is the simplest and leads to the fewest errors.
1.1.Evaluate the brackets:
3(x-2)^2-20=553(x−2)2−20=55
In order to evaluate (x-2)^2(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=553(x−2)(x−2)−20=55
We can evaluate (x-2)(x-2)(x−2)(x−2) using the FOIL method
3(x^2-4x+4)-20=553(x2−4x+4)−20=55
2.2.Expand the brackets:
3(x^2-4x+4)-20=553(x2−4x+4)−20=55
3x^2-12x+12-20=553x2−12x+12−20=55
3.3. Set the equation equal to 00 and collect like terms:
3x^2-12x+12-20=553x2−12x+12−20=55
3x^2-12x+12-20-55=03x2−12x+12−20−55=0
3x^2-12x-63=03x2−12x−63=0
4.4.Solve for the unknown:
3x^2-12x-63=03x2−12x−63=0
The easiest way to solve this is to have a coefficient of 11 for x^2x2, as this will make factorising the equation easier:
x^2-4x-21=0x2−4x−21=0
At this stage, we have a simplified equation equal to 00. In order to find the roots of xx, we need to find two factors which multiply to give cc (in this case -21−21) and add to give bb (in this case -4−4):
x^2-7x+3x-21=0x2−7x+3x−21=0
Then we factorise each 'pair' of terms in the equation:
x(x-7)+3(x-7)=0x(x−7)+3(x−7)=0
Since we are multiplying (x-7)(x−7) by xx and by +3+3, we can put these two multipliers in one bracket to make solving the equation easier:
(x-7)(x+3)=0(x−7)(x+3)=0
5.5.Set each bracket equal to 00 to get values for xx:
Since we know that if a*b=0a⋅b=0, then aa or bb must be equal to 00, we can apply that rule here. But since we don't know which factor equals 00, we set both equal to 00 and solve for both.
x-7=0x−7=0
x=7x=7
x+3=0x+3=0
x=-3x=−3
x_1 = 7x1=7
x_2 = -3x2=−3
