If the set of all a in mathbbR - \1\$a \in \mathbb{R} - \{1\}$, for which the roots of the equation (1 - a)x^2 + 2(a - 3)x + 9 = 0$(1 - a)x^2 + 2(a - 3)x + 9 = 0$ are positive is (-infty, -alpha] cup [beta, gamma)$(-\infty, -\alpha] \cup [\beta, \gamma)$, then 2alpha + beta + gamma$2\alpha + \beta + \gamma$ is equal to ____________.

### Related Formula Since alpha$\alpha$ is a root of x^2 - x + 2 = 0$x^2 - x + 2 = 0$, it must satisfy the equation exactly: alpha^2 - alpha + 2 = 0 Rightarrow alpha^2 = alpha - 2$$\alpha^2 - \alpha + 2 = 0 \Rightarrow \alpha^2 = \alpha - 2$$ This is an essential root reduction property allowing polynomials of high degrees to be collapsed linearly. ### Core Logic We need to evaluate the expression E = alpha^6 + alpha^4 + beta^4 - 5alpha^2$E = \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$. Use the substitution alpha^2 = alpha - 2$\alpha^2 = \alpha - 2$ to iteratively depress the powers of alpha$\alpha$. alpha^4 = (alpha^2)^2 = (alpha - 2)^2 = alpha^2 - 4alpha + 4$$\alpha^4 = (\alpha^2)^2 = (\alpha - 2)^2 = \alpha^2 - 4\alpha + 4$$ Substitute alpha^2 = alpha - 2$\alpha^2 = \alpha - 2$ again into the result: alpha^4 = (alpha - 2) - 4alpha + 4 = -3alpha + 2$$\alpha^4 = (\alpha - 2) - 4\alpha + 4 = -3\alpha + 2$$ Now, generate alpha^6$\alpha^6$ using alpha^4$\alpha^4$: alpha^6 = alpha^4 cdot alpha^2 = (-3alpha + 2)(alpha - 2) = -3alpha^2 + 6alpha + 2alpha - 4 = -3alpha^2 + 8alpha - 4$$\alpha^6 = \alpha^4 \cdot \alpha^2 = (-3\alpha + 2)(\alpha - 2) = -3\alpha^2 + 6\alpha + 2\alpha - 4 = -3\alpha^2 + 8\alpha - 4$$ Substitute alpha^2 = alpha - 2$\alpha^2 = \alpha - 2$ into the result again: alpha^6 = -3(alpha - 2) + 8alpha - 4 = -3alpha + 6 + 8alpha - 4 = 5alpha + 2$$\alpha^6 = -3(\alpha - 2) + 8\alpha - 4 = -3\alpha + 6 + 8\alpha - 4 = 5\alpha + 2$$ ### Step 1: Simplify the Full Expression The symmetry of the roots dictates that beta^4$\beta^4$ behaves identically to alpha^4$\alpha^4$. Thus: beta^4 = -3beta + 2$$\beta^4 = -3\beta + 2$$ Substitute all depressed linear forms back into E = alpha^6 + alpha^4 + beta^4 - 5alpha^2$E = \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$: E = (5alpha + 2) + (-3alpha + 2) + (-3beta + 2) - 5(alpha - 2)$$E = (5\alpha + 2) + (-3\alpha + 2) + (-3\beta + 2) - 5(\alpha - 2)$$ E = 5alpha - 3alpha - 5alpha - 3beta + 2 + 2 + 2 + 10$$E = 5\alpha - 3\alpha - 5\alpha - 3\beta + 2 + 2 + 2 + 10$$ E = -3alpha - 3beta + 16$$E = -3\alpha - 3\beta + 16$$ E = -3(alpha + beta) + 16$$E = -3(\alpha + \beta) + 16$$ ### Step 2: Apply Sum of Roots From the original quadratic equation x^2 - x + 2 = 0$x^2 - x + 2 = 0$, the sum of roots is: alpha + beta = -frac-11 = 1$$\alpha + \beta = -\frac{-1}{1} = 1$$ Substitute this back: E = -3(1) + 16 = 13$$E = -3(1) + 16 = 13$$ (Note: The condition Im(alpha) gt Im(beta)$Im(\alpha) \gt Im(\beta)$ was a distractor since the expression simplified perfectly symmetrically into alpha + beta$\alpha + \beta$ without needing the individual complex values of the roots). ### Pattern Recognition Never compute De Moivre polar forms for high root powers unless the quadratic has roots like omega$\omega$ or i$i$. Always use the characteristic quadratic relation alpha^2 = palpha + q$\alpha^2 = p\alpha + q$ to rapidly step down degrees until everything is strictly linear. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations

### Related Formula z^3-1 = (z-1)(z^2+z+1) quad textCube roots of unity: 1, omega, omega^2$$z^3-1 = (z-1)(z^2+z+1) \quad \text{Cube roots of unity: } 1, \omega, \omega^2$$ omega^3 = 1 quad textand quad 1+omega+omega^2 = 0$$\omega^3 = 1 \quad \text{and} \quad 1+\omega+\omega^2 = 0$$ ### Core Logic Let's first find the roots of the lower degree polynomial: z^3 + 2z^2 + 2z + 1 = 0$$z^3 + 2z^2 + 2z + 1 = 0$$ Group the terms: (z^3 + 1) + 2z(z + 1) = 0$$(z^3 + 1) + 2z(z + 1) = 0$$ (z + 1)(z^2 - z + 1) + 2z(z + 1) = 0$$(z + 1)(z^2 - z + 1) + 2z(z + 1) = 0$$ (z + 1)(z^2 - z + 1 + 2z) = 0$$(z + 1)(z^2 - z + 1 + 2z) = 0$$ (z + 1)(z^2 + z + 1) = 0$$(z + 1)(z^2 + z + 1) = 0$$ Thus, the roots are z = -1$z = -1$, and the roots of z^2 + z + 1 = 0$z^2 + z + 1 = 0$ which are z = omega, omega^2$z = \omega, \omega^2$. ### Step 1: Check z = -1 Substitute z = -1$z = -1$ into the first equation z^1985 + z^100 + 1 = 0$z^{1985} + z^{100} + 1 = 0$: (-1)^1985 + (-1)^100 + 1 = -1 + 1 + 1 = 1 neq 0$$(-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 \neq 0$$ So, z = -1$z = -1$ is not a common root. ### Step 2: Check z = ω and ω² Substitute z = omega$z = \omega$: omega^1985 + omega^100 + 1$$\omega^{1985} + \omega^{100} + 1$$ Reduce the powers modulo 3 (since omega^3 = 1$\omega^3 = 1$): 1985 = 3 times 661 + 2 Rightarrow omega^1985 = omega^2$1985 = 3 \times 661 + 2 \Rightarrow \omega^{1985} = \omega^2$ 100 = 3 times 33 + 1 Rightarrow omega^100 = omega^1 = omega$100 = 3 \times 33 + 1 \Rightarrow \omega^{100} = \omega^1 = \omega$ So, omega^1985 + omega^100 + 1 = omega^2 + omega + 1 = 0$\omega^{1985} + \omega^{100} + 1 = \omega^2 + \omega + 1 = 0$. z = omega$z = \omega$ is a common root. Substitute z = omega^2$z = \omega^2$: (omega^2)^1985 + (omega^2)^100 + 1 = omega^3970 + omega^200 + 1$$(\omega^2)^{1985} + (\omega^2)^{100} + 1 = \omega^{3970} + \omega^{200} + 1$$ 3970 = 3 times 1323 + 1 Rightarrow omega^3970 = omega$3970 = 3 \times 1323 + 1 \Rightarrow \omega^{3970} = \omega$ 200 = 3 times 66 + 2 Rightarrow omega^200 = omega^2$200 = 3 \times 66 + 2 \Rightarrow \omega^{200} = \omega^2$ So, omega^3970 + omega^200 + 1 = omega + omega^2 + 1 = 0$\omega^{3970} + \omega^{200} + 1 = \omega + \omega^2 + 1 = 0$. z = omega^2$z = \omega^2$ is also a common root. ### Step 3: Conclusion There are exactly 2 common roots: omega$\omega$ and omega^2$\omega^2$. ### Pattern Recognition Whenever z^2+z+1$z^2+z+1$ emerges as a factor, its roots omega$\omega$ and omega^2$\omega^2$ can be directly tested in massive degree polynomials using exponent reduction modulo 3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers

### Related Formula |z^n| = |z|^n$|z^n| = |z|^n$ |barz| = |z|$|\bar{z}| = |z|$ ### Core Logic Given the equation: z^2 = -ibarz$z^2 = -i\bar{z}$ Taking the modulus on both sides: |z^2| = |-ibarz|$$|z^2| = |-i\bar{z}|$$ Using properties of modulus: |z|^2 = |-i| cdot |barz|$$|z|^2 = |-i| \cdot |\bar{z}|$$ Since |-i| = 1$|-i| = 1$ and |barz| = |z|$|\bar{z}| = |z|$: |z|^2 = |z|$|z|^2 = |z|$ |z|^2 - |z| = 0$|z|^2 - |z| = 0$ |z|(|z| - 1) = 0$$|z|(|z| - 1) = 0$$ ### Step 1: Applying the non-zero condition This gives two possibilities: |z| = 0$|z| = 0$ or |z| = 1$|z| = 1$. Since z = x + iy$z = x + iy$ and xy neq 0$xy \neq 0$, neither x$x$ nor y$y$ is zero, which implies |z| neq 0$|z| \neq 0$. Therefore, |z| = 1$|z| = 1$. We are asked for |z^2|$|z^2|$: |z^2| = |z|^2 = 1^2 = 1$$|z^2| = |z|^2 = 1^2 = 1$$ ### Pattern Recognition When dealing with equations involving z$z$ and barz$\bar{z}$, applying modulus to both sides rapidly simplifies the problem, turning complex equations into simple real algebraic equations in |z|$|z|$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations

### Related Formula textSum of roots (alpha + beta) = -fracba$$\text{Sum of roots } (\alpha + \beta) = -\frac{b}{a}$$ textProduct of roots (alpha beta) = fracca$$\text{Product of roots } (\alpha \beta) = \frac{c}{a}$$ ### Core Logic For the equation x^2 - 70x + lambda = 0$x^2 - 70x + \lambda = 0$: alpha + beta = 70$\alpha + \beta = 70$ alpha beta = lambda$\alpha \beta = \lambda$ Since alpha, beta in mathbbN$\alpha, \beta \in \mathbb{N}$, their sum is 70. This gives alpha(70 - alpha) = lambda$\alpha(70 - \alpha) = \lambda$. We are given lambda / 2 notin mathbbN$\lambda / 2 \notin \mathbb{N}$ and lambda / 3 notin mathbbN$\lambda / 3 \notin \mathbb{N}$. This means lambda$\lambda$ is not divisible by 2 or 3. So lambda$\lambda$ must be an odd number and not a multiple of 3. Thus, neither alpha$\alpha$ nor beta$\beta$ can be a multiple of 2 or 3. ### Step 1: Finding minimum lambda We need to find the minimum value of lambda$\lambda$. lambda = alpha(70 - alpha)$\lambda = \alpha(70 - \alpha)$. This parabola opens downwards, so the minimum product occurs when the integers alpha$\alpha$ and beta$\beta$ are as far apart as possible. Let's test small values for alpha$\alpha$: If alpha = 1$\alpha = 1$, beta = 69 Rightarrow lambda = 69$\beta = 69 \Rightarrow \lambda = 69$, but 69 = 3 times 23$69 = 3 \times 23$, which is divisible by 3. Rejected. If alpha = 2$\alpha = 2$, divisible by 2. Rejected. If alpha = 3$\alpha = 3$, divisible by 3. Rejected. If alpha = 4$\alpha = 4$, divisible by 2. Rejected. If alpha = 5$\alpha = 5$, beta = 65 Rightarrow lambda = 325$\beta = 65 \Rightarrow \lambda = 325$. Check divisibility: 325 is odd (not div by 2). 3+2+5 = 10$3+2+5 = 10$ (not div by 3). So minimum lambda = 325$\lambda = 325$ with roots alpha = 5, beta = 65$\alpha = 5, \beta = 65$. ### Step 2: Evaluating the target expression We need to compute: E = frac(sqrtalpha - 1 + sqrtbeta - 1)(lambda + 35)|alpha - beta|$$E = \frac{(\sqrt{\alpha - 1} + \sqrt{\beta - 1})(\lambda + 35)}{|\alpha - \beta|}$$ Substitute the values alpha = 5$\alpha = 5$, beta = 65$\beta = 65$, lambda = 325$\lambda = 325$: |alpha - beta| = |5 - 65| = 60$$|\alpha - \beta| = |5 - 65| = 60$$ sqrtalpha - 1 = sqrt4 = 2$$\sqrt{\alpha - 1} = \sqrt{4} = 2$$ sqrtbeta - 1 = sqrt64 = 8$$\sqrt{\beta - 1} = \sqrt{64} = 8$$ E = frac(2 + 8)(325 + 35)60 = frac(10)(360)60 = 10 times 6 = 60$$E = \frac{(2 + 8)(325 + 35)}{60} = \frac{(10)(360)}{60} = 10 \times 6 = 60$$ The result is exactly 60. ### Pattern Recognition Minimizing the product of two numbers with a fixed sum requires them to be as far apart as possible. Divisibility constraints are quickly verified using prime modulus filters (%2 neq 0$%2 \neq 0$, %3 neq 0$%3 \neq 0$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations

### Related Formula a^3+b^3 = (a+b)^3 - 3ab(a+b)$$a^3+b^3 = (a+b)^3 - 3ab(a+b)$$ a^2+b^2 = (a+b)^2 - 2ab$$a^2+b^2 = (a+b)^2 - 2ab$$ a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2$$a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2$$ ### Core Logic Given z_1+z_2=5$z_1+z_2=5$ and z_1^3+z_2^3=20+15i$z_1^3+z_2^3=20+15i$. z_1^3+z_2^3 = (z_1+z_2)^3 - 3z_1z_2(z_1+z_2)$$z_1^3+z_2^3 = (z_1+z_2)^3 - 3z_1z_2(z_1+z_2)$$ 20+15i = 125 - 15z_1z_2$$20+15i = 125 - 15z_1z_2$$ 15z_1z_2 = 105 - 15i implies z_1z_2 = 7-i$$15z_1z_2 = 105 - 15i \implies z_1z_2 = 7-i$$ Now, compute z_1^2+z_2^2$z_1^2+z_2^2$: z_1^2+z_2^2 = (z_1+z_2)^2 - 2z_1z_2 = 25 - 2(7-i) = 11+2i$$z_1^2+z_2^2 = (z_1+z_2)^2 - 2z_1z_2 = 25 - 2(7-i) = 11+2i$$ Now, compute z_1^4+z_2^4$z_1^4+z_2^4$: z_1^4+z_2^4 = (z_1^2+z_2^2)^2 - 2(z_1z_2)^2$$z_1^4+z_2^4 = (z_1^2+z_2^2)^2 - 2(z_1z_2)^2$$ = (11+2i)^2 - 2(7-i)^2$$= (11+2i)^2 - 2(7-i)^2$$ = (121 - 4 + 44i) - 2(49 - 1 - 14i)$$= (121 - 4 + 44i) - 2(49 - 1 - 14i)$$ = 117 + 44i - 2(48 - 14i)$$= 117 + 44i - 2(48 - 14i)$$ = 117 + 44i - 96 + 28i = 21 + 72i$$= 117 + 44i - 96 + 28i = 21 + 72i$$ Finally, find magnitude: |z_1^4+z_2^4| = |21+72i| = sqrt21^2 + 72^2 = sqrt441 + 5184 = sqrt5625 = 75$$|z_1^4+z_2^4| = |21+72i| = \sqrt{21^2 + 72^2} = \sqrt{441 + 5184} = \sqrt{5625} = 75$$ ### Pattern Recognition Standard algebraic identities recursively applied. Extract sum and product, then ladder up to higher powers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations