They call it quadratic equation?

Algebra Level 5

Let $\alpha$ and $\beta$ are two roots of the quadratic equation $x^2+px+q=0$ , where $p$ and $q$ are real numbers, and $q \neq 0$ . Now suppose another quadratic equation $x^2+mx+n=0$ with roots $\alpha+\dfrac{1}{\alpha}$ and $\beta+\dfrac{1}{\beta}$ such that $m+n=0$ .

Find the range of $q$ .

q \in\left [1,\frac{4}{3}\right]

q \in \left[-2,\frac{-1}{3}\right]

q \in \left[\frac{-2}{3},\frac{2}{3}\right]

q \in \left[\frac{1}{3},3\right]

q \in \left[\frac{-4}{3},-1\right]

1 solution

Aditya Dhawan
May 24, 2016

$(It\quad is\quad given\quad that\quad \alpha \quad and\quad \beta \quad are\quad the\quad roots\quad of\quad { x }^{ 2 }+px+q=0\\ Thus\quad by\quad vieta's\quad formulae:\\ (a)\alpha +\beta =-p\\ (b)\alpha \beta =q\\ \\ Also\quad let\quad { R }_{ 1 }\quad and\quad { R }_{ 2\quad }be\quad the\quad roots\quad of\quad the\quad second\quad equation\\ (a)\quad { R }_{ 1 }+{ R }_{ 2 }=\alpha +\beta +\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\quad (\alpha +\beta )\left( 1+\frac { 1 }{ \alpha \beta } \right) =-p-\frac { p }{ q } =-m\Rightarrow \boxed { m=p+\frac { p }{ q } } \\ (b)\quad { R }_{ 1 }{ R }_{ 2 }=\quad \alpha \beta +\frac { \beta }{ \alpha } +\frac { \alpha }{ \beta } +\frac { 1 }{ \alpha \beta } =\quad \alpha \beta \quad +\frac { { (\alpha +\beta ) }^{ 2 }-2\alpha \beta }{ \alpha \beta } +\frac { 1 }{ \alpha \beta } =\quad q+\frac { { p }^{ 2 }-2q }{ q } +\frac { 1 }{ q } \Rightarrow \boxed { n=q+\frac { { p }^{ 2 }-2q }{ q } +\frac { 1 }{ q } } \\ Now,\\ m+n=0\quad (Given)\\ \therefore \quad p+\frac { p }{ q } +q+\frac { { p }^{ 2 }-2q }{ q } +\frac { 1 }{ q } =0\Longrightarrow \quad { p }^{ 2 }+p(q+1)+{ (q-1) }^{ 2 }=0\\ Now\quad for\quad p\quad to\quad be\quad real,\quad the\quad discriminant(D)\quad must\quad be\quad non\quad negative\\ \therefore \quad D\ge 0\quad \Leftrightarrow { (q+1) }^{ 2 }-4({ q-1) }^{ 2 }\ge 0\quad \Leftrightarrow \quad (q-3)(3q-1)\le 0\quad \quad \Leftrightarrow \quad q\in \left[ \frac { 1 }{ 3 } ,3 \right] )$

Moderator note:

Good approach.

Thus far, you have found a necessary condition. You should also explain why this is a sufficient condition, by being clear that we have two-sided implications.

I did not understand the last 3rd step.

Khushi Mehta - 4 years, 6 months ago

Please explain

Khushi Mehta - 4 years, 6 months ago

@Calvin Lin sir I didn't get what you mentioned in the note. Can you explain ?

Aditya Sky - 4 years ago

Do you understand the difference between a necessary condition and a sufficinet condition?

If you read through his solution, it says that "If $q$ satisfies this condition, then $q \in [ \frac{1}{3} , 3 ]$ ." That is a necessary condition.
It does not explain why "Given any $q \in [ \frac{ 1}{3} ,3 ]$ , it will satisfy the conditions.

As an example, if we were asked to "find the range of $y = x^ 2 + (x+1)^2$ , we could say that $x^ 2 ( x + 1) ^2 \geq 0 + 0 = 0$ , and might be tempted to conclude that $y \in [ 0 , \infty )$ . However, this is only a necessary condition, but not a sufficient condition. In particular, the range of $y$ is not $[0, \infty )$ .

Calvin Lin Staff - 4 years ago

I was aware of if then implication but didn't understand as to how it applies to this problem. Isn't it clear that for any $q\,\in\,\left[\frac{1}{3},3\right]$ , the given conditions will be satisfied as the equation from which the range of $q$ follows $\left((q-3)(3q-1)\,=\,0\right)$ is itself derived from the given conditions ?

I've often seen you mentioning about necessary and sufficient conditions. Can you explain when to check an obtained condition is necessary and sufficient.

Aditya Sky - 4 years ago

@Aditya Sky – It is clear to you when you look back at it and start to fill in the gaps.

However, it is not written down in the solution. Unfortunately, despite technological advances, not everyone can read what is not written.

See my previous comment about checking.

Calvin Lin Staff - 4 years ago

Would it be sufficient had the last implication ( for the inequality) been double sided?

Since that would ensure that " For a real p {which satisfies the relation between p and q } to exist , the discriminant must be non-negative. Thereby, we obtain a bound for q. It's also evident that for each value of : $q \in [ \frac{1}{3} , 3 ]$ , the discriminant would be non-negative and thus there would exist a corresponding p."

Aditya Dhawan - 4 years ago

@Aditya Dhawan – Just checking the last condition isn't sufficient. You need to check that all implications to be double sided. $P \Leftarrow Q \Leftrightarrow R$ doesn't give us $P \Leftrightarrow R$ .

IE Suppose you had such a $q$ , the last condition tells us that there exists such a $p$ . But now, how do we know that such a $m$ and $n$ exist?

(Yes, in this case, it's obvious, since $m,n$ are uniquely determined by $p,q$ . But if say $m = p + n^2, n^2 = q + m$ , the we might not be guaranteed a solution. )

Note: This is especially important in geometry problems that were converted into algebra. Failure to check that "lengths are positive" or "triangle inequality is satisfied" can result in a "diagram does not exist" scenario.

Calvin Lin Staff - 4 years ago

They call it quadratic equation?

1 solution

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