In solving the equation (x – 1)(x – 2) = 30, Eric stated that the solution would be
x – 1 = 30 = x = 31
or
(x – 2) = 30 = x = 32
However, at least one of these solutions fails to work when substituted back into the original equation. Why is that? Please help Eric to understand better; solve the problem yourself, and explain your reasoning.

Question

In solving the equation (x – 1)(x – 2) = 30, Eric stated that the solution would be
x – 1 = 30 => x = 31
or
(x – 2) = 30 => x = 32
However, at least one of these solutions fails to work when substituted back into the original equation. Why is that? Please help Eric to understand better; solve the problem yourself, and explain your reasoning.

anonymous · Answer

he forgot that he is applying the zero product rule

anonymous · Answer

check my answer

anonymous · Answer

(x - 1)(x - 2) = 30
 
x² - 3x + 2 = 30
 
x² - 3x - 28 = 0
 
Factor the left side,
 
(x - 7)(x + 4) = 0
 
Now we get x = 7 or x = -4 as the roots.
 
If x = 7, x - 1 = 6, and x - 2 = 5. 6(5) does equal 30
 
If x = -4, x - 1 = -5, and x - 2 = -6. (-5)(-6) also equals 30, so both solutions work

anonymous · Answer

before he solve , he must put the equation in this form ax^2+bx+c=0

anonymous · Answer

nice job :)

anonymous · Answer

very nice
also eric is a moron

anonymous · Answer

(x – 1)(x – 2) = 30 ==> x^2+-3x+2-30 = 0 ==> x^2-3x-28=0 ===> (x-4)(x-7) = 0 ===> x=4 or x=7