We are given a polynomial equation \( P(x) = 4x^4 + 8x^3 - 17x^2 - 12x + 9 = 0 \) with roots \( x_1, x_2, x_3, x_4 \). We need to find the value of \( m \) from the given relation \( \left(4 + x_1^2\right)\left(4 + x_2^2\right)\left(4 + x_3^2\right)\left(4 + x_4^2\right) = \frac{125}{16} m \).
If a polynomial \( P(x) \) of degree \( n \) has a leading coefficient \( a_n \) and roots \( x_1, x_2, \ldots, x_n \), it can be written in factored form as:
\[ P(x) = a_n (x - x_1)(x - x_2) \cdots (x - x_n) \]The product \( \prod_{i=1}^{n} (k^2 + x_i^2) \) can be evaluated by considering the polynomial at complex values. Specifically, we use the identity \( k^2 + x_i^2 = -( -k^2 - x_i^2 ) = -( (ik)^2 - x_i^2 ) = (x_i - ik)(x_i + ik) \). This suggests evaluating \( P(ik) \) and \( P(-ik) \).
Since the coefficients of \( P(x) \) are real, we have \( P(\bar{z}) = \overline{P(z)} \). Therefore, \( P(-ik) = \overline{P(ik)} \), and their product is \( P(ik)P(-ik) = |P(ik)|^2 \).
Step 1: Write the polynomial in its factored form.
The given polynomial is \( P(x) = 4x^4 + 8x^3 - 17x^2 - 12x + 9 \). Since its roots are \( x_1, x_2, x_3, x_4 \), we can write:
\[ P(x) = 4(x - x_1)(x - x_2)(x - x_3)(x - x_4) \]Step 2: Evaluate the polynomial at \( x = 2i \) and \( x = -2i \).
Substituting \( x = 2i \):
\[ P(2i) = 4(2i - x_1)(2i - x_2)(2i - x_3)(2i - x_4) \]Substituting \( x = -2i \):
\[ P(-2i) = 4(-2i - x_1)(-2i - x_2)(-2i - x_3)(-2i - x_4) \]Step 3: Multiply \( P(2i) \) and \( P(-2i) \) to form the desired expression.
\[ P(2i) P(-2i) = 16 \prod_{k=1}^{4} (2i - x_k)(-2i - x_k) \]For each term in the product, we have:
\[ (2i - x_k)(-2i - x_k) = (-x_k + 2i)(-x_k - 2i) = (-x_k)^2 - (2i)^2 = x_k^2 - (-4) = x_k^2 + 4 \]Therefore, the product becomes:
\[ P(2i) P(-2i) = 16 \left(x_1^2 + 4\right)\left(x_2^2 + 4\right)\left(x_3^2 + 4\right)\left(x_4^2 + 4\right) \]From this, we can express the desired quantity as:
\[ \left(4 + x_1^2\right)\left(4 + x_2^2\right)\left(4 + x_3^2\right)\left(4 + x_4^2\right) = \frac{P(2i)P(-2i)}{16} \]Step 4: Calculate the value of \( P(2i) \).
\[ P(2i) = 4(2i)^4 + 8(2i)^3 - 17(2i)^2 - 12(2i) + 9 \]We use the powers of \( i \): \( i^2 = -1, i^3 = -i, i^4 = 1 \).
\[ P(2i) = 4(16 i^4) + 8(8 i^3) - 17(4 i^2) - 24i + 9 \] \[ P(2i) = 4(16) + 8(-8i) - 17(-4) - 24i + 9 \] \[ P(2i) = 64 - 64i + 68 - 24i + 9 \]Combining the real and imaginary parts:
\[ P(2i) = (64 + 68 + 9) + (-64 - 24)i = 141 - 88i \]Step 5: Calculate \( P(2i)P(-2i) \). Since the coefficients of \( P(x) \) are real, \( P(-2i) = \overline{P(2i)} = 141 + 88i \).
Thus, \( P(2i)P(-2i) = |P(2i)|^2 \).
\[ |P(2i)|^2 = (141)^2 + (-88)^2 \] \[ (141)^2 = 19881 \] \[ (-88)^2 = 7744 \] \[ |P(2i)|^2 = 19881 + 7744 = 27625 \]Step 6: Substitute this value back into the expression from Step 3.
\[ \left(4 + x_1^2\right)\left(4 + x_2^2\right)\left(4 + x_3^2\right)\left(4 + x_4^2\right) = \frac{27625}{16} \]Step 7: Equate this result to the given expression to find \( m \).
\[ \frac{27625}{16} = \frac{125}{16} m \]Canceling \( \frac{1}{16} \) from both sides, we get:
\[ 27625 = 125 m \]Solving for \( m \):
\[ m = \frac{27625}{125} \] \[ m = 221 \]Therefore, the value of \( m \) is 221.
The given polynomial can be expressed as:
\[ 4x^4 + 8x^3 - 17x^2 - 12x + 9 = 4(x - x_1)(x - x_2)(x - x_3)(x - x_4). \]
Let \(x_1 = 2i\) and \(x_2 = -2i\). Substituting these values:
\[ 64 - 64i + 68 - 24i + 9 = 4(2i - x_1)(2i - x_2)(2i - x_3)(2i - x_4). \]
Simplify:
\[ 141 - 88i \quad \dots \quad (1) \]
Similarly, for \(-2i\):
\[ 64 + 64i + 68 + 24i + 9 = 4(-2i - x_1)(-2i - x_2)(-2i - x_3)(-2i - x_4). \]
Simplify:
\[ 141 + 88i \quad \dots \quad (2) \]
Using the given condition:
\[ \frac{125}{16}m = \frac{141^2 + 88^2}{16}. \]
Calculate:
\[ m = 221. \]
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,