Dtjytyk

Why does Carol not get rid of the Kree symbol on her suit when she changes its colours?

Why does this expression simplify as such?

What kind of floor tile is this?

What is the difference between lands and mana?

What features enable the Su-25 Frogfoot to operate with such a wide variety of fuels?

Is it necessary to use pronouns with the verb "essere"?

Is this toilet slogan correct usage of the English language?

Giving feedback to someone without sounding prejudiced

Is there any evidence that Cleopatra and Caesarion considered fleeing to India to escape the Romans?

Did the UK lift the requirement for registering SIM cards?

What is Cash Advance APR?

Why is the "ls" command showing permissions of files in a FAT32 partition?

Microchip documentation does not label CAN buss pins on micro controller pinout diagram

How to draw a matrix with arrows in limited space

Does Doodling or Improvising on the Piano Have Any Benefits?

What's the name of the logical fallacy where a debater extends a statement far beyond the original statement to make it true?

Pre-mixing cryogenic fuels and using only one fuel tank

Delete multiple columns using awk or sed

Does the reader need to like the PoV character?

Why is the Sun approximated as a black body at ~ 5800 K?

The Digit Triangles

"It doesn't matter" or "it won't matter"?

How to convince somebody that he is fit for something else, but not this job?

Do we have to expect a queue for the shuttle from Watford Junction to Harry Potter Studio?

If $alpha$, $beta$ are the roots of the equation $ax^2 + 3x + 2 = 0,; a

Quadratic Equation relation between rootsQuadratic equation - $alpha$ and $beta$ RootsRoots of $f'(x)=0$, given that $f(x)=0$ has three real roots $alpha,beta,gamma$If $a>b>0$ and $a^3+b^3+27ab=729$ then $ax^2+bx-9=0$ has roots $alpha,beta,(alpha<beta)$. Find the value of $4beta -aalpha$.Quadratic equation with roots $alpha$ and $beta$How to find the equation with roots $alpha/beta$ and $beta/alpha$, given that $alpha ne beta $, $ alpha^2 = 5alpha -3$, $beta^2 = 5beta -3$Let $alpha $, $beta $ are the roots of $3x^2+x+5=0$If $3x^2-6x+p=0$ has roots $alpha$ and $beta$, then find a quadratic with roots $(alpha+beta)/alpha$ and $(alpha+beta)/beta$Find the value of $frac{alpha^2+2alpha+1}{alpha^2+2alpha+b}$+$frac{beta^2+2beta+1}{beta^2+2beta+b}$If $ V_n= alpha^n+beta^n$ and $alpha,beta$ are roots of $x^2+x-1=0$, then $V_n+{V}_{n-3}=2{V}_{n-2}$?

2

$begingroup$

Show that if $alpha$, $beta$ are the roots of the equation $ax^2 + 3x + 2 = 0,; a<0$, then $$dfrac{(alpha^2)}{(beta)}+dfrac{(beta^2)}{(alpha)}> 0$$

I could only figure out two things

first, that $a = frac{-(3alpha + 2)}{(alpha^2)}$

and $frac{(alpha^2)}{(beta)}$ + $frac{(beta^2)}{(alpha)}$ = $frac{(alpha + beta)(alpha^2 + beta^2 - alphabeta)}{(alphabeta)}$.

please help further

quadratics

share|cite|improve this question

edited Mar 13 at 6:33

Yadati Kiran

2,1121622

asked Mar 13 at 6:18

Virad GuptaVirad Gupta

374

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are you trying to achieve here? Please update your post and include a question.
  $endgroup$
  – maxmilgram
  Mar 13 at 6:29
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Show that if $alpha$, $beta$ are the roots of the equation $ax^2 + 3x + 2 = 0,; a<0$, then $$dfrac{(alpha^2)}{(beta)}+dfrac{(beta^2)}{(alpha)}> 0$$

I could only figure out two things

first, that $a = frac{-(3alpha + 2)}{(alpha^2)}$

and $frac{(alpha^2)}{(beta)}$ + $frac{(beta^2)}{(alpha)}$ = $frac{(alpha + beta)(alpha^2 + beta^2 - alphabeta)}{(alphabeta)}$.

please help further

quadratics

share|cite|improve this question

edited Mar 13 at 6:33

Yadati Kiran

2,1121622

asked Mar 13 at 6:18

Virad GuptaVirad Gupta

374

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are you trying to achieve here? Please update your post and include a question.
  $endgroup$
  – maxmilgram
  Mar 13 at 6:29
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Show that if $alpha$, $beta$ are the roots of the equation $ax^2 + 3x + 2 = 0,; a<0$, then $$dfrac{(alpha^2)}{(beta)}+dfrac{(beta^2)}{(alpha)}> 0$$

I could only figure out two things

first, that $a = frac{-(3alpha + 2)}{(alpha^2)}$

and $frac{(alpha^2)}{(beta)}$ + $frac{(beta^2)}{(alpha)}$ = $frac{(alpha + beta)(alpha^2 + beta^2 - alphabeta)}{(alphabeta)}$.

please help further

quadratics

share|cite|improve this question

edited Mar 13 at 6:33

Yadati Kiran

2,1121622

asked Mar 13 at 6:18

Virad GuptaVirad Gupta

374

$endgroup$

share|cite|improve this question

edited Mar 13 at 6:33

Yadati Kiran

2,1121622

asked Mar 13 at 6:18

Virad GuptaVirad Gupta

374

share|cite|improve this question

edited Mar 13 at 6:33

Yadati Kiran

2,1121622

asked Mar 13 at 6:18

Virad GuptaVirad Gupta

374

edited Mar 13 at 6:33

Yadati Kiran

2,1121622

edited Mar 13 at 6:33

Yadati Kiran

2,1121622

asked Mar 13 at 6:18

Virad GuptaVirad Gupta

374

asked Mar 13 at 6:18

Virad GuptaVirad Gupta

374

3 Answers
3

active

oldest

votes

1

$begingroup$

Hint

The usual method would be to use the formula $$alpha text{ or }beta=frac{-3pmsqrt{9-8a}}{2a}$$ and then work with this which might be rather boring.

Alternatively, you could observe that $$frac{alpha^2}{beta}+frac{beta^2}{alpha}=frac{alpha^3+beta^3}{alpha·beta}$$

You can compute the numerator with Newton's Sums method...

Using Vieta's Formulas might also be a nice approach ;)

share|cite|improve this answer

answered Mar 13 at 6:29

Dr. MathvaDr. Mathva

2,857527

$endgroup$

add a comment | 

1

$begingroup$

I don't think the claim is true.

From your last equation, we continue:

Since $alpha+beta=-3/a>0$ for $a<0,$ we only need bother to show that $$frac{alpha^2+beta^2-alphabeta}{alphabeta}$$ is positive. Thus, we get $$frac{alpha^2+beta^2-2alphabeta+alphabeta}{alphabeta}=frac{(alpha-beta)^2+alphabeta}{alphabeta}=frac{(alpha-beta)^2}{alphabeta} + 1.$$ But we have that $$(alpha-beta)^2=alpha^2+beta^2-2alphabeta+2alphabeta-2alphabeta=(alpha+beta)^2-4alphabeta.$$ Therefore we get that $$frac{alpha^2+beta^2-alphabeta}{alphabeta}=frac{(alpha+beta)^2-3alphabeta}{alphabeta}.$$ Substituting $-3/a$ for the sum, and $2/a,$ the product in RHS of last equation gives $$frac{(-3/a)^2-3(2/a)}{2/a}.$$ Clearly the numerator is positive, but the denominator isn't.

share|cite|improve this answer

edited Mar 13 at 7:01

answered Mar 13 at 6:32

AllawonderAllawonder

2,255616

$endgroup$

add a comment | 

0

$begingroup$

We can write $alphabeta=frac2a$ and $alpha+beta=-frac3a$ by Viète's formulas. Then, with a little help from the binomial theorem,
$$frac{alpha^2}beta+frac{beta^2}alpha=frac{-3/a+((-3/a)^2-3(2/a))}{2/a}$$
$$=frac{9/a-9}2=frac92left(frac1a-1right)$$
Since $a<0$, so is $frac1a-1$, implying that the original expression in $alpha$ and $beta$ is also negative.

share|cite|improve this answer

answered Mar 13 at 6:28

Parcly TaxelParcly Taxel

44.7k1376109

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3146166%2fif-alpha-beta-are-the-roots-of-the-equation-ax2-3x-2-0-a0%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Hint

The usual method would be to use the formula $$alpha text{ or }beta=frac{-3pmsqrt{9-8a}}{2a}$$ and then work with this which might be rather boring.

Alternatively, you could observe that $$frac{alpha^2}{beta}+frac{beta^2}{alpha}=frac{alpha^3+beta^3}{alpha·beta}$$

You can compute the numerator with Newton's Sums method...

Using Vieta's Formulas might also be a nice approach ;)

share|cite|improve this answer

answered Mar 13 at 6:29

Dr. MathvaDr. Mathva

2,857527

$endgroup$

add a comment | 

1

$begingroup$

Hint

The usual method would be to use the formula $$alpha text{ or }beta=frac{-3pmsqrt{9-8a}}{2a}$$ and then work with this which might be rather boring.

Alternatively, you could observe that $$frac{alpha^2}{beta}+frac{beta^2}{alpha}=frac{alpha^3+beta^3}{alpha·beta}$$

You can compute the numerator with Newton's Sums method...

Using Vieta's Formulas might also be a nice approach ;)

share|cite|improve this answer

answered Mar 13 at 6:29

Dr. MathvaDr. Mathva

2,857527

$endgroup$

add a comment | 

$begingroup$

Hint

The usual method would be to use the formula $$alpha text{ or }beta=frac{-3pmsqrt{9-8a}}{2a}$$ and then work with this which might be rather boring.

Alternatively, you could observe that $$frac{alpha^2}{beta}+frac{beta^2}{alpha}=frac{alpha^3+beta^3}{alpha·beta}$$

You can compute the numerator with Newton's Sums method...

Using Vieta's Formulas might also be a nice approach ;)

share|cite|improve this answer

answered Mar 13 at 6:29

Dr. MathvaDr. Mathva

2,857527

$endgroup$

share|cite|improve this answer

answered Mar 13 at 6:29

Dr. MathvaDr. Mathva

2,857527

answered Mar 13 at 6:29

Dr. MathvaDr. Mathva

2,857527

answered Mar 13 at 6:29

Dr. MathvaDr. Mathva

2,857527

1

$begingroup$

I don't think the claim is true.

From your last equation, we continue:

Since $alpha+beta=-3/a>0$ for $a<0,$ we only need bother to show that $$frac{alpha^2+beta^2-alphabeta}{alphabeta}$$ is positive. Thus, we get $$frac{alpha^2+beta^2-2alphabeta+alphabeta}{alphabeta}=frac{(alpha-beta)^2+alphabeta}{alphabeta}=frac{(alpha-beta)^2}{alphabeta} + 1.$$ But we have that $$(alpha-beta)^2=alpha^2+beta^2-2alphabeta+2alphabeta-2alphabeta=(alpha+beta)^2-4alphabeta.$$ Therefore we get that $$frac{alpha^2+beta^2-alphabeta}{alphabeta}=frac{(alpha+beta)^2-3alphabeta}{alphabeta}.$$ Substituting $-3/a$ for the sum, and $2/a,$ the product in RHS of last equation gives $$frac{(-3/a)^2-3(2/a)}{2/a}.$$ Clearly the numerator is positive, but the denominator isn't.

share|cite|improve this answer

edited Mar 13 at 7:01

answered Mar 13 at 6:32

AllawonderAllawonder

2,255616

$endgroup$

add a comment | 

1

$begingroup$

I don't think the claim is true.

From your last equation, we continue:

Since $alpha+beta=-3/a>0$ for $a<0,$ we only need bother to show that $$frac{alpha^2+beta^2-alphabeta}{alphabeta}$$ is positive. Thus, we get $$frac{alpha^2+beta^2-2alphabeta+alphabeta}{alphabeta}=frac{(alpha-beta)^2+alphabeta}{alphabeta}=frac{(alpha-beta)^2}{alphabeta} + 1.$$ But we have that $$(alpha-beta)^2=alpha^2+beta^2-2alphabeta+2alphabeta-2alphabeta=(alpha+beta)^2-4alphabeta.$$ Therefore we get that $$frac{alpha^2+beta^2-alphabeta}{alphabeta}=frac{(alpha+beta)^2-3alphabeta}{alphabeta}.$$ Substituting $-3/a$ for the sum, and $2/a,$ the product in RHS of last equation gives $$frac{(-3/a)^2-3(2/a)}{2/a}.$$ Clearly the numerator is positive, but the denominator isn't.

share|cite|improve this answer

edited Mar 13 at 7:01

answered Mar 13 at 6:32

AllawonderAllawonder

2,255616

$endgroup$

add a comment | 

$begingroup$

I don't think the claim is true.

From your last equation, we continue:

Since $alpha+beta=-3/a>0$ for $a<0,$ we only need bother to show that $$frac{alpha^2+beta^2-alphabeta}{alphabeta}$$ is positive. Thus, we get $$frac{alpha^2+beta^2-2alphabeta+alphabeta}{alphabeta}=frac{(alpha-beta)^2+alphabeta}{alphabeta}=frac{(alpha-beta)^2}{alphabeta} + 1.$$ But we have that $$(alpha-beta)^2=alpha^2+beta^2-2alphabeta+2alphabeta-2alphabeta=(alpha+beta)^2-4alphabeta.$$ Therefore we get that $$frac{alpha^2+beta^2-alphabeta}{alphabeta}=frac{(alpha+beta)^2-3alphabeta}{alphabeta}.$$ Substituting $-3/a$ for the sum, and $2/a,$ the product in RHS of last equation gives $$frac{(-3/a)^2-3(2/a)}{2/a}.$$ Clearly the numerator is positive, but the denominator isn't.

share|cite|improve this answer

edited Mar 13 at 7:01

answered Mar 13 at 6:32

AllawonderAllawonder

2,255616

$endgroup$

share|cite|improve this answer

edited Mar 13 at 7:01

answered Mar 13 at 6:32

AllawonderAllawonder

2,255616

answered Mar 13 at 6:32

AllawonderAllawonder

2,255616

answered Mar 13 at 6:32

AllawonderAllawonder

2,255616

0

$begingroup$

We can write $alphabeta=frac2a$ and $alpha+beta=-frac3a$ by Viète's formulas. Then, with a little help from the binomial theorem,
$$frac{alpha^2}beta+frac{beta^2}alpha=frac{-3/a+((-3/a)^2-3(2/a))}{2/a}$$
$$=frac{9/a-9}2=frac92left(frac1a-1right)$$
Since $a<0$, so is $frac1a-1$, implying that the original expression in $alpha$ and $beta$ is also negative.

share|cite|improve this answer

answered Mar 13 at 6:28

Parcly TaxelParcly Taxel

44.7k1376109

$endgroup$

add a comment | 

0

$begingroup$

We can write $alphabeta=frac2a$ and $alpha+beta=-frac3a$ by Viète's formulas. Then, with a little help from the binomial theorem,
$$frac{alpha^2}beta+frac{beta^2}alpha=frac{-3/a+((-3/a)^2-3(2/a))}{2/a}$$
$$=frac{9/a-9}2=frac92left(frac1a-1right)$$
Since $a<0$, so is $frac1a-1$, implying that the original expression in $alpha$ and $beta$ is also negative.

share|cite|improve this answer

answered Mar 13 at 6:28

Parcly TaxelParcly Taxel

44.7k1376109

$endgroup$

add a comment | 

$begingroup$

We can write $alphabeta=frac2a$ and $alpha+beta=-frac3a$ by Viète's formulas. Then, with a little help from the binomial theorem,
$$frac{alpha^2}beta+frac{beta^2}alpha=frac{-3/a+((-3/a)^2-3(2/a))}{2/a}$$
$$=frac{9/a-9}2=frac92left(frac1a-1right)$$
Since $a<0$, so is $frac1a-1$, implying that the original expression in $alpha$ and $beta$ is also negative.

share|cite|improve this answer

answered Mar 13 at 6:28

Parcly TaxelParcly Taxel

44.7k1376109

$endgroup$

share|cite|improve this answer

answered Mar 13 at 6:28

Parcly TaxelParcly Taxel

44.7k1376109

answered Mar 13 at 6:28

Parcly TaxelParcly Taxel

44.7k1376109

answered Mar 13 at 6:28

Parcly TaxelParcly Taxel

44.7k1376109