Quadratic equation with a parameter

Question

The Problem:

What is the value of $q$ in $x^2+6x+q=0$, if the distance between solutions is greater than 4.

The Result: $q\in(-\infty,5)$

My Solution:

$$x_{1}+x_{2}=-6\\x_{1}\cdot x_{2}=q\\|x_{1}-x_{2}|>4 $$ $$ x_{1}-x_{2} >4\hspace{0.2cm} \diagup^{2} \hspace{1cm} \lor \hspace{1cm}x_{1}-x_{2}<-4\hspace{0.2cm} \diagup^{2}\\(x_{1}+x_{2})^2 >16+4x_{1} x_{2}\hspace{0.2cm} \hspace{1cm} \hspace{1cm}(x_{1}+x_{2})^2 <16+4x_{1} x_{2}\hspace{0.2cm}\\ (-6)^2>16+4q\hspace{3cm} (-6)^2<16+4q\\ 5>q\hspace{6.5cm} 5<q \\ q\in(-\infty,5)\cup(5,\infty) $$ Squareing the inequalities helped me substitute Vieta's formulas, does this cause a problem of getting wrong solutions? If so, why? What is the other way of solving this problem using Vieta's formulas?

Answer 1

i have got $$|2\sqrt{9-q}|>4$$ and from here we get $$\sqrt{9-q}>2$$ and from here we get $$5>q$$

Answer 2

We need $|-3+\sqrt{9-q}-(-3-\sqrt{9-q})|>4$ or $$\sqrt{9-q}>2$$ or $$9-q>4,$$ which gives $$q<5.$$

The mistake is that $$|x_1-x_2|>4\Leftrightarrow(x_1-x_2)^2>16,$$ which gives $$(x_1+x_2)^2-4x_1x_2>16$$ or $$36-4q>16,$$ which gives $q<5.$

Answer 3

(1) $0< a< b$ implies $0< a^2< b^2$.

(2) And $b< a< 0$ implies $0< -a< -b$ which also implies $0< a^2< b^2$.

Thus, what is wrong is the second column of your calculation. The correct calculation should be:

$$x_{1}-x_{2}<-4\hspace{0.2cm} \diagup^{2}\\ (x_{1}-x_{2})^2>(-4)^2\\ (x_{1}+x_{2})^2 >16+4x_{1} x_{2}\hspace{0.2cm}\\ \vdots$$

which lead us to the same result obtained in the first column. (Note that we are applying (2) with $a=-4$ and $b=x_1-x_2$.)

Alternatively, you could start the second column as follows:

$$x_{1}-x_{2}<-4\\ x_{2}-x_{1}>4\\ x_{2}-x_{1}>4\hspace{0.2cm} \diagup^{2}\\ (x_{2}-x_{1})^2>16\\ \vdots$$

Answer 4

The main error lies in squaring an inequality when there are negative numbers involved.

For example, $-8<4$ is true but if you blindly square both sides, you get $64<16$ which is false.

If you know both sides are positive, then it is valid to square both sides, but this can introduce extra solutions. For example, $x>2 \Longrightarrow x^2>4 \Longleftrightarrow x>2 \vee x<-2$. The squaring step goes only in one direction. This is because the squaring step can be broken down as follows: $$x>2 \Longleftrightarrow\\ x-2>0 \Longrightarrow\\ (x-2)(x+2)>0 \Longleftrightarrow\\ x^2-4>0 \Longleftrightarrow\\ x^2>4$$ There is one step which is a multiplication by $x+2$, and that step is only valid if $x+2$ is positive because otherwise the inequality would have to be flipped around.