# Is this root extraneous?

### Is this root extraneous?

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

$$(x^{2}-5x+6)\sqrt{x^{2}+7}=0$$

Here's the solution:

$$x^{2}-5x+6=0; \sqrt{ x^{2}-7 }=0$$
$$D=5^{2}-4*6; ( \sqrt{x^{2}-7} )^{2}=0^{2}$$
$$D=25-24;x^{2}-7=0$$
$$D=1;(x - sqrt{7} )( x+\sqrt{7} )=0$$
$$x=\frac{ 5+1 }{2}; x=\frac{ 5-1 }{2};x-\sqrt{7}=0;x+\sqrt{7}=0$$
$$x=\frac{6}{2};x=\frac{4}{2};x=\sqrt{7}; x=-\sqrt{7}$$
$$x=3;x=2$$

Let's check the solution:

$$( (-\sqrt{7})^{2}-5(-\sqrt{7})+6 )\sqrt{ (-\sqrt{7})^{2}-7 }=( 7+5\sqrt{7}+6 )\sqrt{ 7-7 }=( 13+5\sqrt{7} )\sqrt{0}=( 13+5\sqrt{7} )*0=0$$
$$( 2^{2}-5*2+6 )\sqrt{ 2^{2}-7 }=( 4-10+6 )\sqrt{ 4-7 }=0*\sqrt{ -3 }=0*i*\sqrt{3}=0$$
$$( ( \sqrt{7} )^{2}-5\sqrt{7}+6)\sqrt{ (\sqrt{7})^2-7 }=( 7-5\sqrt{7}+6 )\sqrt{ 7-7 }=( 13-5\sqrt{7} )\sqrt{0}=( 13-5\sqrt{7} )*0=0$$
$$( 3^{2}-5*3+6 )\sqrt{ 3^{2}-7 }=( 9-15+6 )\sqrt{ 9-7 }=0*\sqrt{2}=0$$

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

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 $$0*\sqrt{-3}=0*i\sqrt{3}$$. 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?
Guest

### Re: Is this root extraneous?

I would say that the question is not well written. The left hand side of the equation $$(x^2-5x+6)\sqrt{x^2-7}=0$$ is undefined in the region $$-\sqrt{7}<x< \sqrt{7}$$, it is perfectly acceptable (if I really wanted to) to define square roots of negative numbers to always be 0 instead of imaginary numbers. The question should explicitly state the region of $$x$$ the solutions should lie in (or how square roots of negative numbers are to be treated), or say something like $$x$$ must be a real value for which the left hand side of the equation is well defined.

Having said that, I would probably agree with your teacher. In general even if it is unstated I would only consider those $$x$$ for which the left hand side is clearly well defined (assuming you are not supposed to know about complex numbers).

Probably not the answer you wanted to hear, and I'm sure there will be lots of people that will disagree with me, but that's my opinion for what it's worth. Don't let it discourage you, you may lose marks due to semantics, but your maths is fine, be happy in the knowledge that you understand complex numbers and so know things the rest of your class don't.

Hope this helped,

R. Baber.
Guest

### Re: Is this root extraneous?

Today I have asked my Math teacher. She said that, firstly, complex roots are different from real roots, and so $$\sqrt{-3}$$ is undefined, and secondly, in school we shouldn't use complex numbers anywhere, even if thery are multiplied to 0, and in exam using complex numbers will be considered mistake.
Guest

### Re: Is this root extraneous?

nice post

yasir7719

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