quadratic equation with zero product

Question

I have this equation

${3x^2 -12 = -9}$

The answer the text book gives is ${x = 1}$ or ${x = 3}$.

But I would solve it by first of all dividing 3 on both sides which gives:

${x^2 -4 = -3}$

Then add +3 to both sides which gives:

${x^2 -1}$

Which gives ${x = \pm1}$

I've obviously taken a wrong turn.

Answer 1

If your equation will be like this $$3x^{ 2 }-12x=-9$$ then it has two roots such as $$\\ x^{ 2 }-4x+3=0\\ \left( x-1 \right) \left( x-3 \right) =0\\ x_{ 1 }=1,{ x }_{ 2 }=3$$,obviously there a typo in the book

Answer 2

It's likely that the intended equation was $3x^2 -12x = -9$ which does have the given roots.

With the typo, your solution is correct.

Answer 3

If you have this equation

${3\ x^2 -12 x = -9},$

the answer the text book gives ${x = 1}$ or ${x = 3},$ is right.

The $x$ before $=$ got missed, a typo.