How to solve quadratic equation with parameter for solutions in some range

Question

Given an equation: $2x^2 - (p+3)x + p + 1 = 0$ and I have to find two real solutions and these solutions both have to be in range $(-1,3)$.

So assumptions are: $a \neq 0 \land \Delta > 0 \land x_1 \in (-1, 3) \land x_2 \in(-1,3) $

Firstly, $a=2\neq 0$, so $p \in R$

Secondly, $ \Delta = (p+3)^2 - 4*2*(p+1) = p^2+6p+9-8p-8=p^2-2p+1=(p-1)^2$

So we calcuate when the $\Delta > 0$

$(p-1)^2 > 0$

So $p \in (-\infty, 1) \cup (1, +\infty)$

Now, we need to calculate when the solutions are in range I mentioned. And that's where the problem occurs. I tried to just do it by solving $-1<x_1<3 \land -1<x_2<3$. But see it:

$x_1=(p+3-(p-1))/4 = 4/4=1$

$x_2=(p+3+p+1)/4=(2p+4)/4 = (1/2)p + 1$

The first double inequality is valid for all real $p$. So solve the second:

$-1 < (1/2)p+1 \land (1/2)p+1<3$

$p>-4 \land p<4$

And here I know I did something bad, because the valid answer (for whole problem) is $p\in(-3,1) \cup (1,5)$, while the inequality I solved above restricts the result to $4$ at max.

I heard that I shall use Vieta's formulas. But imagine a example where we search for solutions greater than two. The one man said me in that case I should do $x_1+x_2>4$ and $x_1*x_2 > 4$. Even I did one task using it and it worked, but I completely don't understand why. Because it doesn't exclude case like $x_1 = 5 \land x_2=1$, which of course should not be valid solutions.

So how can I solve these these types of equations?

Answer 1

I'm going to first point out an error in the method mentioned in question, and then show how to solve it using Vieta's formulas.

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There is an error in $x_2=(p+3+p\color{red}+1)/4$.

It should be $x_2=(p+3+p\color{red}-1)/4$.

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To solve it using Vieta's formulas, we use the following claim :

Claim : $a\gt 0$ and $b\gt 0$ if and only if $a+b\gt 0$ and $ab\gt 0$.

Proof of the claim :

• If $a\gt 0$ and $b\gt 0$, then $a+b\gt 0$ and $ab\gt 0$.

• If $a+b\gt 0$ and $ab\gt 0$, then it follows that $a,b$ have the same sign. Suppose that $a\lt 0$ and $b\lt 0$. Then, $a+b\lt 0$ which contradicts $a+b\gt 0$. So, $a\gt 0$ and $b\gt 0$.$\ \square$

I heard that I shall use Vieta's formulas. But imagine a example where we search for solutions greater than two. The one man said me in that case I should do $x_1+x_2>4$ and $x_1∗x_2>4$.

It is not correct that $x_1+x_2>4$ and $x_1∗x_2>4$.

Using the claim, we have $$\begin{align}&x_1>2,x_2>2 \\\\&\iff x_1-2>0,x_2-2>0 \\\\&\iff (x_1-2)+(x_2-2)\gt 0,(x_1-2)(x_2-2)>0 \\\\&\iff (x_1+x_2)-4\gt 0, x_1x_2-2(x_1+x_2)+4\gt 0\end{align}$$

Our question can be solved simlarly.

Using the claim, we have $$\begin{align}&-1\lt x_1\lt 3,-1\lt x_2\lt 3 \\\\&\iff 3-x_1\gt 0,3-x_2\gt 0,x_1+1\gt 0,x_2+1\gt 0 \\\\&\iff (3-x_1)+(3-x_2)\gt 0,(3-x_1)(3-x_2)\gt 0, \\&\qquad (x_1+1)+(x_2+1)\gt 0, (x_1+1)(x_2+1)\gt 0 \\\\&\iff 6-(x_1+x_2)\gt 0,9-3(x_1+x_2)+x_1x_2\gt 0, \\&\qquad (x_1+x_2)+2\gt 0, x_1x_2+(x_1+x_2)+1\gt 0 \\\\&\iff 6-\frac{p+3}{2}\gt 0, 9-3\times\frac{p+3}{2}+\frac{p+1}{2}\gt 0, \\&\qquad \frac{p+3}{2}+2\gt 0,\frac{p+1}{2}+\frac{p+3}{2}+1\gt 0 \\\\&\iff p\lt 9,p\lt 5,p\gt -7, p\gt -3 \\\\&\iff -3\lt p\lt 5\end{align}$$

You already got $p\not=1$, so the answer is $$\color{red}{p\in (-3,1)\cup (1,5)}$$

Answer 2

I have reached that if a, b and c are functions in terms of p, and we need to make the solutions of the equation $a(p)x^2+b(p)+c(p)=0$ in the interval $(n,m)$, so we need to solve for p the below inequality: $$a(p)n^2+b(p)n<-c<a(p)m^2+b(p)m$$.

First, consider $a(p)>0$, and $\Delta>0$. So we have that $b^2(p)>4a(p)c(p)$. With the quadratic formula, we have that $$x=\frac{-b(p)\pm \sqrt{b^2(p)-4a(p)c(p)}}{2a(p)}$$ So if we need to this solutions be on the interval $(n,m)$, we have that: $$n<\frac{-b(p)\pm \sqrt{b^2(p)-4a(p)c(p)}}{2a(p)}<m$$ $$2a(p)n+b(p)<\pm \sqrt{\Delta} < 2a(p)m+b(p)$$ $$4a^2(p)n^2+4a(p)b(p)n+b^2(p)<b^2(p)-4a(p)c(p)<4a^2(p)m^2+4a(p)b(p)m+b(p)^2$$ $$an^2+bn<-c<am^2+bm$$

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Answer 3

The question and the existing answer are too long, so I decided to write an answer after I read the first paragraph.

The discriminant $D$ of the equation equals $$(p+3)^2-4\cdot 2(p+1)=p^2-2p+1=(p-1)^2,$$ so both roots of the equation are real (but can coincide). The roots of the equation are $\frac14 \left(p+3\pm \sqrt{D}\right)= \frac14 \left(p+3\pm (p-1)\right)$. That is the roots are $\frac {p+1}2$ and $1$. Thus we need to have $\frac {p+1}2\in (-1,3)$, that is $p\in (-3,5)$. The roots coincide when $\frac {p+1}2=1$, that is when $p=1$.

Answer 4

A vigilant student would have immediately noticed that $x_1=1$ is a root of the equation $$2x^2-(p+3)x+p+1=0$$ and by polynomial division factorize the equation like $$(x-1)(2x-p-1)=0.$$ Then the other root is $x_2=\frac{p+1}2$. With the given conditions: $-1<x_2<3$ and $x_2\neq x_1$ (If "Two real solutions" mean they are different. The valid answer in OP suggests so.) he would have $$p\in(-3,1)\cup(1,5).$$ But İ followed the long way, I completed the square like $$2\left(x-\frac{p+3}4\right)^2=\frac{(p+3)^2}8-p-1$$ $$\left(x-\frac{p+3}4\right)^2=\left(\frac{p-1}4\right)^2$$ $$\left|x-\frac{p+3}4\right|=\frac{p-1}4$$ and hence the roots are $$x_1=-\frac{p-1}4+\frac{p+3}4=1,$$ $$x_2=\frac{p-1}4+\frac{p+3}4=\frac{p+1}2.$$ Then $$-1<\frac{p+1}2<3\iff$$ $$-2<p+1<6\iff$$ $$\bbox[yellow,5px,border:2px solid blue]{-3<p<5}$$ and $$\frac{p+1}2\neq1\iff\bbox[yellow,5px,border:2px solid blue]{p\neq1}.$$