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Question: If \(x,\mspace{6mu} y,\mspace{6mu} z\), then the roots of the equation \(u = x^{2} + 4y^{2} + 9z^{2...

If x,6muy,6muzx,\mspace{6mu} y,\mspace{6mu} z, then the roots of the equation

u=x2+4y2+9z26yz3zxzxyu = x^{2} + 4y^{2} + 9z^{2} - 6yz - 3zx - zxy are.

A

Equal

B

Imaginary

C

Real

D

None of these

Answer

Real

Explanation

Solution

We have a[2,8]a \in \lbrack 2,8\rbrack Let roots are a[2,8]a \in \lbrack - 2,8\rbrack and a(2,8)a \in (2,8)

Let a(2,8)p,q{1,2,3,4}a \in ( - 2,8)p,q \in \{ 1,2,3,4\}

Given that, px2+qx+1=0px^{2} + qx + 1 = 0

Putting this value, we get

x2(3k1)x+x^{2} - (3k - 1)x +.

Hence roots are real.