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Quadratic Function Confusion (1 Viewer)

Real Madrid

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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

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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

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Real Madrid said:
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

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

bubblesss

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Real Madrid said:
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

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lyounamu said:
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? :p

Sezenator said:
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

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

tommykins

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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

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

Both the roots are equal.
 

tommykins

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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
 

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