Question

If x, y, and z are real variables satisfying the equations x + y + z = 5 and xy + yz+ zx = 8. Determine the range of x.

Answer 1

Answer:

Explanation:

Since, x + y + z = 5 (Given)

Therefore, z = 5 – (y + x) . . . . . . . . . . . . (1)

On Substituting the values of Equation (1) in xy + yz+ zx = 8 we get,

xy + z (y+ z) = 8,

i.e. xy + (y + z) (5 – y- x) = 8,

Or, xy + 5y – y2 – yx + 5x – xy – x2 = 8

Or, y2 – (5 – x)y – 5x + 8 + x2 = 0

Now, b2 – 4ac (D) ≥ 0 [Since, y is real]

i.e. (5 – x)2 – 4 (x2 – 5x + 8) ≥ 0,

Or, x2 + 25 – 10x – 4x2 + 20x – 32 ≥ 0,

Or, – 3x2 + 10x – 7 ≥ 0,

Or, 3x2 – 10x + 7 ≤ 0 . . . . . . . . . . . (2)

Now, the roots of quadratic equation 3x2 – 10x + 7 = 0

(x-1) (x-7/3) $\geq$ 0

x ∈ [1, 7/3]

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