polynomial equations

[carebear]

Find the quadratic whose roots are -1 and 1/3 and whose value at x = 2 is 10.

I found the equation to be p(x) = 2/3(x+1)(3x-1), which I then multiplied to get 2x^2 +4/3x-2/3 so I multiplied the entire equation by 3 to get 6x^2 + 4x - 2. However, when I check x = 2 into that equation, I do not get 10, I get 30. What am I doing wrong?

What is the correct answer?

 

[Mrspi]

Find the quadratic whose roots are -1 and 1/3 and whose value at x = 2 is 10.

I found the equation to be p(x) = 2/3(x+1)(3x-1), which I then multiplied to get 2x^2 +4/3x-2/3 so I multiplied the entire equation by 3 to get 6x^2 + 4x - 2. However, when I check x = 2 into that equation, I do not get 10, I get 30. What am I doing wrong?

What is the correct answer?

Ok...you had p(x) = 2x[sup:v4nw6ki3]2[/sup:v4nw6ki3] + (4/3)x - (2/3)

You want the result to be 10 when you substitute 2 for x. Let's see what happens if we DO that:

p(x) = 2x[sup:v4nw6ki3]2[/sup:v4nw6ki3] + (4/3)x - (2/3)
p(2) = 2(2)[sup:v4nw6ki3]2[/sup:v4nw6ki3] + (4/3)(2) - (2/3)
p(2) = 2(4) + (8/3) - (2/3)
p(2) = 8 + (8/3) - (2/3)
8 + (6/3)
p(2) =8 + 2
p(2) = 10

It works! When you use 2 for x, you get a value of 10 for p(x)....

Now, you multiplied by 3 to get rid of the fractions, I assume. Did you remember that you need to multiply BOTH SIDES of the equation by 3?

In other words, if you have
p(x) = 2x[sup:v4nw6ki3]2[/sup:v4nw6ki3] + (4/3)x - (2/3)
and you multiply by 3, you need to multiply BOTH SIDES by 3:

3*p(x) = 3*[2x[sup:v4nw6ki3]2[/sup:v4nw6ki3] + (4/3)x - (2/3)]
3*p(x) = 6x[sup:v4nw6ki3]2[/sup:v4nw6ki3] + 4x - 2
And when you substitute 2 for x on the right side, you will, in fact, get 30....but that is (according to what is on the left side) THREE TIMES as large as the value of p(2).

 

[soroban]

Hello, carebear!

\(\displaystyle \text{Find the quadratic whose roots are }-1\text{ and }\tfrac{1}{3}\text{ and whose value at }x = 2\text{ is }10.\)

\(\displaystyle \text{I found the equation to be: }\(x) \:=\: \tfrac{2}{3}(x+1)(3x-1)\)

\(\displaystyle \text{which I then multiplied to get: }\(x) \:=\:2x^2 + \tfrac{4}{3}x - \tfrac{2}{3}\)

\(\displaystyle \text{so I multiplied the entire equation by 3 to get: }\: 6x^2 + 4x - 2\)

Why? . . . Now you have THREE TIMES the desired quadratic!


\(\displaystyle \text{However, when I check }x = 2\text{ into that equation, I do not get }10\text{, I get }30.\)
. . . Well, duh!

 

[carebear]

I multiplied it by 3 to get rid of the fraction....the denominator of 3. I thought the left side of the equation was 0 so multiplied it by 3 and still got 0. Judging by your very blunt response, it is obviously wrong.

 

[mmm4444bot]

I thought the left side of the equation was 0

What equation?

You did fine, until you started adding unnecessary steps after you already had a correct answer:

(2/3)(x + 1)(3x - 1)

That is a quadratic polynomial.

You have some terminology issues to resolve. In your original post, you refer to an expression as an equation. Equations always contain an equal sign; expressions never do. Until the distinction between a polynomial and a polynomial equation is clear in your head, you'll confuse yourself.

Also, the word "root" belongs to a polynomial, not an equation. (That's a clue.) Functions do not have roots; equations do not have roots. Polynomials have roots; functions have zeros, and equations have solutions.

This exercise asks for "the quadratic whose roots are -1 and 1/3" (emphasis mine). The word "roots" tells us that the answer is a polynomial, even though the instructions do not. I mean, it would be clearer to request "the quadratic polynomial whose roots are -1 and 1/3", but the instruction is clear enough as given to distinguish between the need to report a polynomial versus some expression equal to zero (i.e., an equation).

If your instructor actually wants a function definition, then the word "roots" is misleading, and the entire class should be confused.

Unless there are additional, unposted instructions, your initial result (factored form) is good enough because it satisfies the three given conditions. If your instructor requires standard form (Ax^2 + Bx + C), then they need to specify that.

You do not need to simplify your result. You do not need to set it equal to zero. You do not need to clear any fractions. In other words, you were done.