Question

I just noticed a common mistake when I was thinking about the answer to the question \(3x^2 = 18x, x = ?\).

\(\textbf{Wrong way:}\)

Divide both sides by \(3\).
\( \color{Black}{\Rightarrow x^2 = 6x }\)

Divide both sides by \(x\).
\(x = 6\)

I admit that it does it for \(6\), but we are missing one solution.

\(\textbf{Correct way:}\)

Form a quadratic equation from the given:
\(\Rightarrow 3x^2 - 18x = 0\)

Divide both sides by \(3\).
\(\Rightarrow x^2 - 6x = 0\)

Factor.
\(\Rightarrow x(x - 6) = 0\)

Use the zero-product rule now.

Answer 1

I hope that this prevented you from committing mistakes in the future.

Answer 2

You know this now??

Answer 3

?

Answer 4

The degree of x is here 2 that is sufficient to tell that there must be two values of x..

Answer 5

Yeah. But that's not exactly what I pointed out here.

Answer 6

Also, \(x^2 + 10x + 25 = 0\) has only one solution, so your statement doesn't apply always.

Answer 7

It has two solutions dear..

The condition is D = 0 here which says that the two solutions are real and equal here..

x = -5 here..

Answer 8

so i cannot just divide by x when ever i feel like it eh,?

Answer 9

In this case - no.

Answer 10

i can always divide by three though ,

Answer 11

Yes you can divide by constant but you don't know the value of x because it is a variable so its values can be 0..

So dividing by 0 is always avoided..

Answer 12

Yep yep

Answer 13

ok, i understand now

Answer 14

Actually Parth, you skipped a step:

Factor out the three first:

\(3x(x - 6) = 0\)

Then, divide by 3:

\(x(x-6)=0\)

Anytime you divide by a variable, you are eliminating a solution set, so it is definitely an algebraic a "no-no".