Hi everyone! Recently I wrote a math test in my school. The last task in that test was to solve this equation:

[tex](x^{2}-5x+6)\sqrt{x^{2}+7}=0[/tex]

Here's the solution:

[tex]x^{2}-5x+6=0; \sqrt{ x^{2}-7 }=0[/tex]

[tex]D=5^{2}-4*6; ( \sqrt{x^{2}-7} )^{2}=0^{2}[/tex]

[tex]D=25-24;x^{2}-7=0[/tex]

[tex]D=1;(x - sqrt{7} )( x+\sqrt{7} )=0[/tex]

[tex]x=\frac{ 5+1 }{2}; x=\frac{ 5-1 }{2};x-\sqrt{7}=0;x+\sqrt{7}=0[/tex]

[tex]x=\frac{6}{2};x=\frac{4}{2};x=\sqrt{7}; x=-\sqrt{7}[/tex]

[tex]x=3;x=2[/tex]

Let's check the solution:

[tex]( (-\sqrt{7})^{2}-5(-\sqrt{7})+6 )\sqrt{ (-\sqrt{7})^{2}-7 }=( 7+5\sqrt{7}+6 )\sqrt{ 7-7 }=( 13+5\sqrt{7} )\sqrt{0}=( 13+5\sqrt{7} )*0=0[/tex]

[tex]( 2^{2}-5*2+6 )\sqrt{ 2^{2}-7 }=( 4-10+6 )\sqrt{ 4-7 }=0*\sqrt{ -3 }=0*i*\sqrt{3}=0[/tex]

[tex]( ( \sqrt{7} )^{2}-5\sqrt{7}+6)\sqrt{ (\sqrt{7})^2-7 }=( 7-5\sqrt{7}+6 )\sqrt{ 7-7 }=( 13-5\sqrt{7} )\sqrt{0}=( 13-5\sqrt{7} )*0=0[/tex]

[tex]( 3^{2}-5*3+6 )\sqrt{ 3^{2}-7 }=( 9-15+6 )\sqrt{ 9-7 }=0*\sqrt{2}=0[/tex]

So, I think that this equation has four roots: [tex]-\sqrt{7}, 2, \sqrt{7}[/tex] and [tex]3[/tex].

But in our school we aren't taught about complex numbers. My teacher said that we'll learn about it in university. And because of this, we are said that square roots of negative numbers don't exist. Look at the root 2. If we substitute "x" in the equation to "2", we'll have [tex]0*\sqrt{-3}=0*i\sqrt{3}[/tex]. Although multiplying complex (imaginary in this occasion) number to 0 gives 0 (so the root is correct), teacher considered that this root was extraneous, and writing it to an answer was considered a mistake.

What do you think?