Question

Question #17784

Answer 1

Let `r =`the first root then `2r` is the second root

To have a leading coefficient of 2, we have the following factors:

`2(x-r)(x-2r)`

Multiply the binomials:

`2(x^2-3r+2r^2)`

Distribute the constant:

`2x^2-6r+4r^2`

Matching coefficients we obtain two equations:

`6r = k" [1]"`
`4r^2 = k" [2]"`

Solve equation [1] for r in terms of k:

`r = k/6" [1.1]"`

Substitute into equation [2]:

`4(k/6)^2 = k`

Simplify the left side:

`k^2/9 = k`

Subtract k from both sides:

`k^2/9 - k=0`

Multiply by 9:

`k^2 - 9k=0`

Factor:

`k(k-9) = 0`

`k = 0 and k = 9`

We are told that `k!=0`, therefore, we discard that root:

`k = 9`

Check the quadratic `k = 9`:

`2x^2-9x+9 = 0`

Factor:

`(2x-3)(x-3) = 0`

`x = 3/2 and x = 3`

The second root is twice the first, therefore, this checks.

Answer 2

`k = 9`

Explanation:

There are different ways to solve this, so this is not necessarily the best.

For quadratic: `ax^2+bx+c`

If the roots of a quadratic are `alpha` and `beta` then:

`alpha+beta= (-b)/a` and `alphaxxbeta=c/a`

Since one root is twice the other, let `alpha` = `alpha` and `beta = 2alpha`

So we have: `3alpha = (-b)/a` and `2alpha^2 = c/a`

From: `2x^2-kx+k` we have:

`3alpha=k/2` and `2alpha^2=k/2`

`3alpha=k/2=>alpha=k/6`

Put `alpha=k/6` in `2alpha^2=k/2`

`2(k/6)^2=k/2=> 2(k^2/36)= k/2=>4k^2-36k=0`

Factoring: `4k^2-36k=0`

`4k(k-9)=0=> k=0` or `k= 9` (discard `0` since from question this is not valid)

Our equation is now:

`2x^2-9x+9=0`

Check:

Factor `2x^2-9x+9`

`(2x-3)(x-3)=0=> x=3` or `x= 3/2`

`2xx3/2=3`

So one root is twice the other: