Fundamental Drill#3
Find the value of α such that:
4 x 2 − 4 x + 6 + ( 3 α + 2 ) y 2 − 6 α x y − 4 y ≤ 0
for x , y satisfying the inequality where x , y ∈ R .
You can try more of my fundamental problems here .
The answer is 4.
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1 solution
Hmm, I think you might want to reword your problem, especially the "for all reals x , y " part, because one might think you meant the inequality 4 x 2 − 4 x + 6 + ( 3 α + 2 ) y 2 − 6 α x y − 4 y ⩽ 0 holds for some value α (which they have to find) no matter what real value x and y they choose, hence when they see this problem, they might try setting y = 0 to get 4 x 2 − 4 x + 6 ⩽ 0 , which is impossible, and thus get stumped.
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So I should add that y = 0 to remove the clouds of doubt is it?
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You could just ask to find a real α such that x , y satisfy the inequality, where x , y ∈ R .
4 x 2 − 4 x + 6 + ( 3 α + 2 ) y 2 − 6 α x y − 4 y ≤ 0
3 ( α y 2 − 2 α x y + x 2 ) + ( x 2 − 4 x + 4 ) + 2 ( y 2 − 2 y + 1 ) ≤ 0
3 ( α y − x ) 2 + ( x − 2 ) 2 + 2 ( y − 1 ) 2 ≤ 0
However,
( α y − x ) 2 ≥ 0
( x − 2 ) 2 ≥ 0
( y − 1 ) 2 ≥ 0
∴ 3 ( α y − x ) 2 + ( x − 2 ) 2 + 2 ( y − 1 ) 2 ≥ 0
There is solution if and only if the equality holds.
When 3 ( α y − x ) 2 + ( x − 2 ) 2 + 2 ( y − 1 ) 2 = 0
⎩ ⎪ ⎨ ⎪ ⎧ α y − x = 0 x − 2 = 0 y − 1 = 0
x = 2 , y = 1
∴ α ⋅ 1 − 2 = 0
α = 2
∴ α = 4