Classic Algebric Problem

Algebra Level 5

IF

x + y + z = 4 x+y+z=4

x 2 + y 2 + z 2 = 6 x^2+y^2+z^2=6

for real numbers x , y x,y and z z . If the exhaustive range of values that x x can take is given by [ γ , δ ] [\gamma ,\delta] , find 12 ( γ + δ ) 12(\gamma +\delta)

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The answer is 32.

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2 solutions

Solving the 2 equations, we get y = 1 2 ( 3 x 2 + 8 x 4 a + 4 ) , z = 1 2 ( 3 x 2 + 8 x 4 a + 4 ) y = \frac{1}{2}(-\sqrt{-3x^{2}+8x-4}-a+4), z= \frac{1}{2}(\sqrt{-3x^{2}+8x-4}-a+4) or y = 1 2 ( 3 x 2 + 8 x 4 a + 4 ) , z = 1 2 ( 3 x 2 + 8 x 4 a + 4 ) y = \frac{1}{2}(\sqrt{-3x^{2}+8x-4}-a+4), z= \frac{1}{2}(-\sqrt{-3x^{2}+8x-4}-a+4)

Hence, in order for y y and z z to be real, we have the condition that 3 x 2 + 8 x 4 a + 4 0 -3x^{2}+8x-4-a+4 \geq 0

Sketching out the graph for 3 x 2 + 8 x 4 a + 4 -3x^{2}+8x-4-a+4 we get 2 3 x 2 \frac{2}{3}\leq x\leq2

Thus, γ = 2 3 , δ = 2 \gamma=\dfrac{2}{3}, \delta=2 , so 12 ( γ + δ ) = 32 12(\gamma +\delta)=\huge 32

Akshay Sharma
Jan 15, 2016

Using Cauchy Schwaz on y,z.

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