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quadratic (1 Viewer)

ta1g

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how would u do this?

1) Find the value of m in x^2 + 2mx -6 = 0 if one of the roots is 2

2) In the quadratic function x^2 + mx + 2 = 0 the roots are consecutive, find the values of m
 

word.

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1. if x = 2 is a root, then f(2) = 0
i.e. f(x) = (x - 2)(x - k) = 0
x2 - kx - 2x + 2k = 0
x2 - x(k + 2) + 2k = 0

2k = -6, so k = -3
x2 + x - 6 = 0
m = 1/2

2. let f(x) = (x - k)(x - {k + 1})
(x - k)(x - k - 1) = 0
x2 - kx - x - kx + k2 + k = 0
x2 - x(2k + 1) + (k2 + k) = 0

k2 + k = 2
k2 + k - 2 = 0
(k - 1)(k + 2) = 0
k = 1, -2

x2 - 3x + 2 = 0; m = -3
x2 + 3x + 2 = 0; m = 3
 

Riviet

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Another way of doing the first one:
Since x=2 (the root)
sub. x=2 into the equation because whenever we sub a root into the equation it satisfies it, we want to find the value of m such that when we sub x=2 in, we get 0
4+4m-6=0
4m=2
.: m=1/2
I think this is approach is more logical and easier to understand

Here's a slight variation on the second using product of roots:
let the roots be k and k+1, then:
k(k+1)=2
k^2+k=2
k^2+k-2=0
(k+2)(k-1)=0
.: k= -2, 1

Now if we use sum of roots then we get this:
k+k+1= -m
sub. values of k in
-4+1=-m
m= 3
1+1+1=-m
m=-3
 
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