# Range of quadratic function using discriminant

Mathematics Asked by Mutse on December 21, 2020

Let $$x^2-2xy-3y^2=4$$. Then find the range of $$2x^2-2xy+y^2$$.

Let $$2x^2-2xy+y^2=a$$.
Then $$ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$$.
We divide both side by $$y^2$$ and let $$t=frac{x}{y}$$.
Then it implies $$(a-8)t^2-(2a-8)t-(3a+4)=0$$.
Since its discriminant is not negative, $$frac{Delta}{4}ge 0implies a^2-7a-4ge 0$$. It gives us $$a$$ can have negative values like $$-1$$. But if clearly contracts $$a-4=x^2+4y^2ge 0implies age 4$$. Where did I mistake?

Others have pointed out where you went wrong. I just want to provide an alternative proof.

$$x^2-2xy-3y^2=4=(x+y)(x-3y)$$

Denote $$z=x+y, w=x-3y$$, then $$zw=4, x=(3z+w)/4, y=(z-w)/4$$.

We want to find the range of $$2x^2-2xy+y^2=frac{1}{16} (5w^2+14wz+13z^2)=frac 72+frac{1}{16} (5w^2+13z^2) tag 1$$ under the constraint $$zw=4$$.

Clearly $$(1)$$ can be as large as you want. And if you apply AM-GM you get the minimum $$frac 72+frac{1}{16} (5w^2+13z^2) ge frac 72 + frac{1}{16} 2 sqrt{5} sqrt{13}|wz|=frac{7+sqrt{65}}{2}$$

Answered by Neat Math on December 21, 2020

Consider the curves $$x^2-2xy-3y^2=4 (C)$$ and $$2x^2-2xy+y^2=k^2+4 (E)$$ At the points where they intersect, we can substitute the value of $$-2xy$$ from the first into the second to get $$x^2+4y^2=k^2 (E)$$ This represents an ellipse centered at the origin. Now, either with a bit of graph sketching, or by considering $$(C)$$ as a quadratic in $$x^2$$, deduce that $$y$$ can range anywhere in the real numbers, and that $$(C)$$ is a hyperbola. Increasing $$k$$ in $$(E)$$ just enlarges the ellipse, and it’s not hard to see that there is no upper bound on $$k$$ for $$(E)$$ and $$(C)$$ to intersect. For obtaining a lower bound, we need to find $$k$$ such that the two curves touch. Note that if $$(x,y)$$ lies on the two curves, then so does $$(-x,-y)$$. So we’re looking for exactly two intersections. Then substituting $$x=pmsqrt{k^2-4y^2}$$ in $$(C)$$ gives a quadratic in $$y^2$$: $$65y^4-y^2(18k^2-56)+(k^2-4)^2 = 0$$ Setting the discriminant equal to zero will give $$k^2=frac{sqrt{65}-1}{2}$$ and hence we have the desired range: $$frac{sqrt{65}+7}{2} le 2x^2-2xy+y^2 =k^2+4lt infty$$

What was wrong in your approach? What you stated was a necessary condition on $$a$$, but that doesn’t mean that $$a$$ could really equal anything in that range.

Answered by Tavish on December 21, 2020

Method$$#1:$$

For real $$t,$$ $$a^2-7a-4ge0$$

If $$x=ty$$

$$4=x^2-2xy-3y^2=y^2(t^2-2t-3)$$

$$t^2-2t-3=dfrac4{y^2}>0$$

$$implies(t-3)(t+1)>0$$

Either $$t<-1$$ or $$t>3$$

So, the values of $$a$$ must satisfy this condition as well

Method$$#2:$$

$$4=(x-y)^2-(2y)^2$$

WLOG $$y=tan t, x-y=2sec timplies x=2sec t+tan t$$

$$a=2(2sec t+tan t)^2-2(2sec t+tan t)tan t+tan^2t$$

Multiplying both sides by $$cos^2t=1-sin^2t,$$

$$(1+a)sin^2t+4sin t+8-a=0$$

What if $$a+1=0?$$

Else
$$sin t=dfrac{-2pmsqrt{a^2-7a-4}}{a+1}$$

As $$sin t$$ is real, the discriminant must be $$ge0$$

But that is not sufficient, we need $$-1ledfrac{-2pmsqrt{a^2-7a-4}}{a+1}le1$$ as well for real $$t$$

Also, $$dfrac xy=2csc t+1$$ $$impliesdfrac xyge2+1$$ or $$dfrac xyle-2+1$$

Answered by lab bhattacharjee on December 21, 2020

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