Quadratic equation and absolute value problems

[Otho]

Hi, I have a problem with these:

1. If the roots of 2x^2-kx+14=0 are integers, then which of the following could be the value of constant k?

a. 6
b. 8
c. 10
d. 12
e. 16

I know that I should use 'sum of the roots' and 'product of the roots', yet I can't solve it. I don't know why an explanation in the book says that sum of the roots (r1+r2= k/2), and the product (r1r2=14/2=7) imply that the possible roots are either {7,1} or {-7,-1}. It also says that it implies that k=16 or -16.

Absolute value problem:

2. If x + sqrt( (1-sqrt(3))^2 ) = 3, then x=. Yes, I know it's gibberish.

a. 1 - sqrt(3)
b. 4 - sqrt(3)
c. 2 + sqrt(3)
d. sqrt(3) -2
e. sqrt(3) -4

My solution:
1) sqrt( (1-sqrt(3))^2 ) = 3 - x
2) |1 - sqrt(3)| = 3 - x At this moment I'm stuck, because I don't want to use calculator in order to keep sqrt(3).

Thank you!

 

[pka]

Absolute value problem:
2. If x + sqrt( (1-sqrt(3))^2 ) = 3, then x=. Yes, I know it's gibberish.

That is not gibberish: \(\displaystyle x+\sqrt{(1-\sqrt{3})^2} =3\)

That is \(\displaystyle x+\sqrt3-1=3\)

 

[Otho]

Using quadratic formula leads to:
x = [k +- SQRT(k^2 - 112)] / 4
Which of the 5 choices makes SQRT(k^2 - 112) = integer?

Ad 2
The only one is answer E, but is it possible to solve this without plugging in?

Does it help if we let the equation be x^2 + (b/a)x + (c/a)=0, where (-b/a) is a sum of the roots, and (c/a) is a product?

Ad 1
x + sqrt(3) - 1 = 3, Why do we change signs of sqrt(3) and 1?

Thanks again to both of you.

 

[stapel]

1. If the roots of 2x^2-kx+14=0 are integers, then which of the following could be the value of constant k?

a. 6, b. 8, c. 10, d. 12, e. 16

I know that I should use 'sum of the roots' and 'product of the roots', yet I can't solve it. I don't know why an explanation in the book says that sum of the roots (r1+r2= k/2), and the product (r1r2=14/2=7) imply that the possible roots are either {7,1} or {-7,-1}. It also says that it implies that k=16 or -16.

The "r1+r2=k/2" thing comes from the sum of the roots, r1 and r2, being equal to half of the coefficient of the linear term. The "r1r2=14/2" thing comes from the product of the roots, r1 and r2, being equal to the quotient of the constant term and leading coefficient. In other words, they got those equations from the formulas that you know you are supposed to use. They just plugged into those formulas.

If the product is 7, what other integer factor pairs could you possibly have?

2. If x + sqrt( (1-sqrt(3))^2 ) = 3, then x=.

My solution:
1) sqrt( (1-sqrt(3))^2 ) = 3 - x
2) |1 - sqrt(3)| = 3 - x At this moment I'm stuck...

Is the square root of 3 more or less than 1? So what is the sign on the subtraction inside the absolute-value bars? So what happens to the expression inside when you take the bars off? So what equation do you end up with? What then is the solution?

 

[HallsofIvy]

The simplest way to do 1 is to note that if a and b are roots then \(\displaystyle 2(x- a)(x- b)= 2x^2- 2(a+ b)x+ ab= 2x^2- kx+ 14\) so we must have ab= 14. The only way the roots can be integers is if a= 1, b= 14, a= 2, b= 7, a=14, b= 1, or a= 7, b=2. If a= 1 and b= 14 or a= 14 and b= 1, then 2(a+ b)= 2(15)= 30. If a= 2 and b= 7 or b= 7 and a= 2, then 2(a+ b)= 2(9)= 8