A Maths - Quadratic Equations and Inequalities
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Show that the solutions of the equation x^2 +kx = 3-k are real for all real values of k.
Here's what I did.
x^2 +kx -3 +k = 0
a = 1, b = k, c = -3+k (AS DERIVED FROM THE FORMULA ax^2 + bx +c)
Discriminant is more than or equal to 0.
D (DISCRIMINANT) = b2 -4ac
=k2 -4(1)(-3+k)
=k2 - 4k +12 >/= 0
k2 - 4k >/= -12
k(k-4) >/= -12
How do I continue from here? Or did I made any error in the earlier statements? Please comment. Thanks.
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x²+kx=3-k
x²+kx+(k-3)=0
Taking discriminant,
b²-4ac
= k²-4(1)(k-3)
= k²-4k+12
= (k²-4k+4)+8
= (k-2)²+8
Since (k-2)² ≥ 0 for all real values of k, the discriminant is always positive. Hence, the solutions to the equation are real for all real values of k. -
Originally posted by bonkysleuth:
...
k(k-4) >/= -12
didn't go through the whole thing, but if in future u get something like that, it's wrong.
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Originally posted by ^tamago^:
x²+kx=3-k
x²+kx+(k-3)=0
Taking discriminant,
b²-4ac
= k²-4(1)(k-3)
= k²-4k+12
= (k²-4k+4)+8
= (k-2)²+8
Since (k-2)² ≥ 0 for all real values of k, the discriminant is always positive. Hence, the solutions to the equation are real for all real values of k.How do you know that (k-2)2 ≥ 0 ? that's the discriminant thingy, i know. but what about the 8 that is inside the equation you've written?
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Originally posted by bonkysleuth:
How do you know that (k-2)² ≥ 0 ? that's the discriminant thingy, i know. but what about the 8 that is inside the equation you've written?
(any real equation/function/expression)² ≥ 0. e.g. (-1)² = 1, (-4)² = 16. dun ask why, at your stage, anything squared must be positive, until you go JC where you deal with unreal/complex numbers.
if (k-2)² ≥ 0, then (k-2)² + 8 ≥ 8, which is definitely positive or > 0.
For your question, you are required to show the roots of an equation are real. You will simply need to prove that its root is always positive (and not zero and/or negative), thus completing the square is the fastest way to achieve this. -
Originally posted by ^tamago^:
x²+kx=3-k
x²+kx+(k-3)=0
Taking discriminant,
b²-4ac
= k²-4(1)(k-3)
= k²-4k+12
= (k²-4k+4)+8
= (k-2)²+8
Since (k-2)² ≥ 0 for all real values of k, the discriminant is always positive. Hence, the solutions to the equation are real for all real values of k.Tamago did a good job here. Now if you notice the discriminant obtained ==> (k-2)²+8 this is a quadratic curve which is always above the x-axis aka y values are always positive so no matter what is the value of k, your discriminant is always > 0 hence the solutions of the equation x^2 +kx = 3-k are real for all real values of k.
Cheers,
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thanks.
