Faith tries to solve the equation x^2-3x=0 by first dividing both sides by x to get x-3=0. She concludes that the only solution is x=3. Is she correct? Explain.

Question

this is factoring trinomials subject please help

Answer 1

The thing is that if Faith divides both sides by x to cancel an x, that changes the equation from a parabolic into a linear equation. She is on the right track in that an x can be factored out of the equation:

x(x - 3) = 0

Since we're multiplying, two numbers, this means either:

x can equal 0, or x - 3 can equal 0. This means there is a second possibility: x = 0, and we can test that in the original equation:

0² - 3*0 = 0

0 - 0 = 0

0 = 0

 

If we tried that in Faith's adjusted equation, we would get:

0 - 3 = 0

-3 ≠ 0

 

Or worse if you think about it a little:

0 - 3 = 0/0

Anything divided by 0 is undefined.

Answer 2

She is correct in that 3 is one of the solutions. By dividing both sides by "x" she eliminates another solution.

This should be solved by factoring.

 

x² - 3x = 0

x(x - 3) = 0

 

x = 0          or

x - 3 = 0 ⇒ x = 3

 

{0, 3}

 

As a side note, the number of solutions to a polynomial equation is equal to the degree of the polynomial.