Equation Help

[SaLoose]

Hello! I am completely stuck on how to answer this question, any help or guidance would be greatly appreciated!

(b) Find the equation whose roots are 1, -2 and 1/2

Thanks!

 

[tkhunny]

"completely stuck" -- I NEVER believe this. You CAN'T have NOTHING unless your teacher is a sadist (which I consider highly unlikely).

If I tell you "x = 1" is a root, can you give me a factor that caused us to know that this is a root?

If I tell you "(x-2)" is a factor, can you give me the root associated with that factor?

 

[topsquark]

You CAN'T have NOTHING unless your teacher is a sadist (which I consider highly unlikely).

Why not? I am!

(b) Find the equation whose roots are 1, -2 and 1/2

I presume that this should be "Find the smallest degree polynomial such that it has zeros at 1, -2, and 1/2.

Hint: Three distinct zeros means what for the degree of the polynomial? Is it 1st, 2nd, 3rd, 4th...

-Dan

 

[pka]

Hello! I am completely stuck on how to answer this question, any help or guidance would be greatly appreciated!
(b) Find the equation whose roots are 1, -2 and 1/2

The answer is: \(\displaystyle (x-1)(x+2)(2x-1)=0\). That is the result of the factor theorem.

 

[Harry_the_cat]

The answer is: \(\displaystyle (x-1)(x+2)(2x-1)=0\). That is the result of the factor theorem.

That is AN equation with the given roots. The general equation would be
\(\displaystyle a(x-1)^b(x+2)^c(2x-1)^d=0\)

 

[Steven G]

Find the equation whose roots are 1, -2 and 1/2. I don't think so. Who makes up these problems?

 

[Harry_the_cat]

Find the equation whose roots are 1, -2 and 1/2. I don't think so. Who makes up these problems?

We are assuming the OP has written it as given.

 

[Otis]

… Who makes up these problems?

SaLoose has posted only part of the exercise (part b). We may be missing given information regarding 'lowest degree' and 'leading coefficient'.

?

 

[DaisySimpson]

After solving we will get the answer (x−1)(x+2)(2x−1)=0.

 

[Otis]

After solving we will get the answer (x−1)(x+2)(2x−1)=0.

Very good, Daisy. Your answer matches the answer shown in post #4.

(SaLoose seems to be taking a break.)

?

 

[Steven G]

That is AN equation with the given roots. The general equation would be
\(\displaystyle a(x-1)^b(x+2)^c(2x-1)^d=0\)

What are your constraints on your constants a, b, c and d??

 

[Harry_the_cat]

What are your constraints on your constants a, b, c and d??

Real where b, c, d are positive.