Quadratic Function Confusion (2 Viewers)

[Real Madrid]

In the quadratic equation...
(k-2)x^2-5x+2k+3=0
The roots are recipricals of each other
Find the value of k.

 

[tommykins]

let roots be a, 1/a

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

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

 

[lyounamu]

In the quadratic equation...
(k-2)x^2-5x+2k+3=0
The roots are recipricals of each other
Find the value of k.

Let the roots be @, 1/@

@ . 1/@ = c/a = (2k+3)/(k-2)
1 = (2k+3)/(k-2)


k-2 = 2k+3
-k = 5
k = -5

WTF... My solution is exactly same as tommykins....WTF

I have never experienced this before.

 

[Sezenator]

omg this sucks any of you can give me pin pointers on how to get back on track with 2u maths?

 

[bubblesss]

In the quadratic equation...
(k-2)x^2-5x+2k+3=0
The roots are recipricals of each other
Find the value of k.

a = k-2 b=-5 c= 2k +3
let the roots be a and 1/a

product of roots is a x 1/a =c/a
1 = 2k +3/k - 2
k-2 = 2k+3
-5 = k

 

[tommykins]

Let the roots be @, 1/@

@ . 1/@ = c/a = (2k+3)/(k-2)
1 = (2k+3)/(k-2)


k-2 = 2k+3
-k = 5
k = -5

WTF... My solution is exactly same as tommykins....WTF

I have never experienced this before.

Looooooooool, must be love?

omg this sucks any of you can give me pin pointers on how to get back on track with 2u maths?

What problems are you having exactly?

 

[Real Madrid]

In the quadratic equation
x^2+mx+2=0
the roots are consecutive
Fins the Values of m.

 

[tommykins]

let roots be a , a+1
a(a+1) = 2
a² + a = 2
a² + a - 2 = 0
(a+2)(a-1) = 0
a= -2 or 1

let a = -2, thus a+1 = -1

-m = a + a + 1 = -2 + -1 = -3
.'.m = 3.

let a = 1, a+1 = 2
-m = a+a+1 = 1+2 = 3
m = -3.

 

[Real Madrid]

Find the values of k in the equation
x^2+(k+1)x+([k+1]/4)=0

Both the roots are equal.

 

[tommykins]

Discriminant ≥ 0

(k+1)² - 4.([k+1]/4) ≥ 0
(k+1)² - k - 1 ≥ 0
k² + 2k + 1 - k - 1 ≥ 0
k² + k ≥ 0
k(k+1) ≥ 0
k ≥ 0 or -1 ≥ k