Math Is Fun Forum

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Help!

Let f(x)=ax^2+bx+a, where a and b are constants and a is not equal to 0.

If one of the roots of the equation f(x)=0 is x=4, what is the other root? Explain your answer.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help!

Hi;

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Help!

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Re: Help!

I said that the other root was -1/4 because I first factored ax^2+bx+a into (ax+1)(x+a).  Since one of the roots were 4, I knew that a had to be -4.  Then, substituting -4 for a in ax+1=0, I got that it was -1/4.

Last edited by dazzle1230 (2016-11-02 09:22:21)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help!

I do not think the other root is - 1 / 4.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Help!

I said that the other root was -1/4 because I first factored ax^2+bx+a into (ax+1)(x+a).  Since one of the roots were 4, I knew that a had to be -4.  Then, substituting -4 for a in ax+1=0, I got that it was -1/4.

(ax+1)(x+a)=0. As per this breakup.-a and -1/a are the two roots. Which of them is 4 ?

Last edited by thickhead (2016-11-02 16:51:37)

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

dazzle1230

Member

Registered: 2016-05-17

Posts: 92

Re: Help!

I see now.  It's 1/4

evene

Member

Registered: 2015-10-18

Posts: 272

Re: Help!

Another way, is to notice that the coefficients go in a,b,a.

Dividing both sides by x, we arrive at

And isolating x, we get

And by Vieta's formula, we see that the other root must be 1/4.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Help!

In ax^2+bx+c=0 product of roots =c/a
Here it is 1. If one root is 4 the other is 1/4. Why complicated logic?

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

evene

Member

Registered: 2015-10-18

Posts: 272

Re: Help!

Complicated logic? I don't understand.