Quadratic Function Question (1 Viewer)

[Blackmancan]

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

I keep getting the wrong answer :/

 

[_deloso]

make the two roots theta and theta+1 since it is consecutive. then you do theta=-b/a and theta+1=c/a and solve simultaneously

 

[hscishard]

Let roots be a and a+1
Addition of roots = 2a + 1 = 5(n-1)/2
Product of roots = a(a+1) = 6

Solve equation for the 2nd equation. The substitute both values of a into the first one to find n. There are two n values

 

[hscishard]

I got -1 and 3, you can check if you got it right by substituting these n vales in the original equation and solve it to check if the roots are consecutive

 

[nazfiz]

roots= α and (α+1) α= alpha
product of roots= c/a= 12/2=6
therefore:

α(α+1)=6
α^2+α-6=0
(α+3)(α-2)=0
α= -3,2

sum of roots= -b/a=-3+2=-1
therefore
5(n+1)/2= -1
5n-5=-2
n=3/5


was that the answer?

Edit: I might be wrong sorry

 

[AAEldar]

I got -1 and 3

D:

Need to check my answer then...

Lololol mine aren't right at all, but I have no idea what I've done wrong hahahaha.

Ignore me, I suck.

 

[Parvee]

The textbook says n=-1,3
I've seen that questions before :L

 

[Blackmancan]

I got -1 and 3, you can check if you got it right by substituting these n vales in the original equation and solve it to check if the roots are consecutive

Yeah thats the right answer

 

[hscishard]

roots= α and (α+1) α= alpha
product of roots= c/a= 12/2=6
therefore:

α(α+1)=6
α^2+α-6=0
(α+3)(α-2)=0
α= -3,2

sum of roots= -b/a=-3+2=-1
therefore
5(n+1)/2= -1
5n-5=-2
n=3/5


was that the answer?

Edit: I might be wrong sorry

You got the -3 and 2 part. You gotta remember that a+1 is the other root for each case. I.e. Sum of roots = -3+(-3+1) and sum of roots = 2+(2+1)

 

[bleakarcher]

i scribbled the working on paint lol, its n=-1, 3

 

[Blackmancan]

Hey i got another question om stuck on:
Show that the roots of the quadratic 3mx^2 +2x+3m=0 are always reciprocals of one another.

 

[D94]

Show that the roots of the quadratic 3mx^2 +2x+3m=0 are always reciprocals of one another.

Let the roots be A and B
Product of Roots: A*B = c/a = 3m/3m = 1
So AB = 1 therefore A = 1/B and B = 1/A.

 

[nazfiz]

You got the -3 and 2 part. You gotta remember that a+1 is the other root for each case. I.e. Sum of roots = -3+(-3+1) and sum of roots = 2+(2+1)

Ohkay. I see what I did wrong. Thanks