quadratic help urgent (1 Viewer)

[ta1g]

1) (k-2)x^2 - 5x +2k + 3 =0 , the roots are reciprocals of eachother. Find the value of k.

(the answer should be k= -5)

2) Find values of n in the equation 2x^2-5(n-1)x + 12=0 if the two roots are consecutive.

(the answer should be n= -1, 3)

If anyone could show me simply how to work these out it would be great, thanks.

 

[Dreamerish*~]

1. Let the roots be α and β.

α + β = -b/a = 5/(k - 2)

αβ = c/a = (2k + 3)/(k - 2)

Since the roots are reciprocals of each other, α = 1/β.

. : 1/β x β = (2k + 3)/(k - 2)

1 = (2k + 3)/(k - 2)

k - 2 = 2k + 3

k = -5.

 

[Riviet]

Damn, i wish 4u was as easy as this :
Let the roots be p and 1/p
then 1 + 1/p = 5/(k-2) [using sum of roots rule] (1)
and p x 1/p = (2k+3)/(k-2) (using product of roots rule] (2)
From (2), k-2 = 2k+3
.: k= -5

2nd one needs a little more working:
let the roots be p and p+1 since they are consecutive
then p+p+1 = 5(n-1)/2 [using sum of roots rule]
and p(p+1) = 6 [using product of roots rule]
p^2+p-6=0
(p+3)(p-2)=0
p=-3, 2

Now we sub in values of p in other equation:
-6+1=(5n-5)/2
-10=5n-5
.: n=-1

4+1=(5n-5)/2
10=5n-5
.: n=3

P.S I personally think using one variable makes it easier to understand and do the question instead of using 2 but either way you will still get the same answer

 

[Dreamerish*~]

2) Find values of n in the equation 2x^2-5(n-1)x + 12=0 if the two roots are consecutive.

2. Let the roots be α and β.

α + β = -b/a = 5(n - 1)/2

αβ = c/a = 12/2 = 6

Since the two roots are consecutive, β = α + 1.

α + (α + 1) = 5(n - 1)/2

α(α + 1) = 6

α² + α - 6 = 0

(α + 3)(α - 2) = 0

. : α = 2, a = -3

α + (α + 1) = 5(n - 1)/2

When α = 2,

2 + 2 + 1 = 5(n - 1)/2

10 = 5n - 5

15 = 5n

. : n = 3

When α = -3

-3 - 3 + 1 = 5(n - 1)/2

-10 = 5n - 5

-5 = 5n

. : n = 1