P_1^{}(x) = ax^2-bx-c, P_2^{}(x) = bx^2-cx-a, P_3(x) = cx^2-ax-b be three quadritic polynomials . Suppose there exists a real number α such that P_1 (α) = P_2(α) = P_3(α). Prove that a =b= c

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Q. Let

P_{1} (x) = a x^{2} - b x - c, P_{2} (x) = b x^{2} - c x - a, P_{3} (x) = c x^{2} - a x - b

be three quadratic polynomials where

a, b, c

are non-zero real numbers. Suppose there exists a real number

α

such that

P_{1} (α) = P_{2} (α) = P_{3} (α)

. Prove that

a = b = c

Q. If roots of the equation

a x^{2} + b x + c = 0

are

α, β

, then the equation whose roots are

\frac{1 + α}{1 - α}, \frac{1 + β}{1 - β},

where

α \neq 1, β \neq 1

Q. Suppose

a, b, c

are three distinct real numbers and

p (x)

is a real quadratic polynomial such that

⎡ ⎢ ⎣ \begin{matrix} 4 a^{2} & 4 a & 1 4 b^{2} & 4 b & 1 4 c^{2} & 4 c & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} p (- 1) p (1) p (1) \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 3 a^{2} + 3 a 3 b^{2} + 3 b 3 c^{2} + 3 c \end{matrix} ⎤ ⎥ ⎦

Given $a x^{2} + b x + c \geq 0, b x^{2} + c x + a \geq 0, c x^{2} + a x + b \geq 0$ where $a \neq b \neq c$ and $a, b, c ϵ R$ . Now
$\frac{a^{2} + b^{2} + c^{2}}{a b + b c + c a}$ cannot take the value(s)

Q. Let a,b,c be any real numbers. Suppose that there are real numbers

x, y, z

not all zero such that

x = c y + b z, y = a z + c x

and

z = b x + a y .

Then

a^{2} + b^{2} + c^{2} + 2 a b c

is equal to

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