If the set of all a in mathbbR$a \in \mathbb{R}$, for which the equation 2x^2 + (a - 5)x + 15 = 3a$2x^2 + (a - 5)x + 15 = 3a$ has no real root, is the interval (alpha, beta)$(\alpha, \beta)$, and X = \x in mathbbZ : alpha < x < beta\$X = \{x \in \mathbb{Z} : \alpha < x < \beta\}$, then sum_x in X x^2$\sum_{x \in X} x^2$ is equal to

### Related Formula Modulus definition rule: |x| = begincases x, & x ge 0 \\ -x, & x < 0 endcases$$|x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}$$ ### Core Logic Analyze the roots by splitting into cases around the critical threshold x = frac32$x = \frac{3}{2}$: **Case I:** x ge frac32$x \ge \frac{3}{2}$ x^2 + 2x - 3 - 4 = 0 implies x^2 + 2x - 7 = 0 implies x = 2sqrt2 - 1$$x^2 + 2x - 3 - 4 = 0 \implies x^2 + 2x - 7 = 0 \implies x = 2\sqrt{2} - 1$$ (We select the positive root since 2sqrt2-1 ge 1.5$2\sqrt{2}-1 \ge 1.5$). ### Step 1: Evaluating the alternate domain branch **Case II:** x < frac32$x < \frac{3}{2}$ x^2 - (2x - 3) - 4 = 0 implies x^2 - 2x - 1 = 0 implies x = 1 - sqrt2$$x^2 - (2x - 3) - 4 = 0 \implies x^2 - 2x - 1 = 0 \implies x = 1 - \sqrt{2}$$ (We select 1-sqrt2$1-\sqrt{2}$ since it satisfies the inequality constraint). ### Step 2: Summing the Squares of the Roots textSum of Squares = (2sqrt2 - 1)^2 + (1 - sqrt2)^2$$\text{Sum of Squares} = (2\sqrt{2} - 1)^2 + (1 - \sqrt{2})^2$$ = (8 - 4sqrt2 + 1) + (1 - 2sqrt2 + 2) = 12 - 6sqrt2 = 6(2 - sqrt2)$$= (8 - 4\sqrt{2} + 1) + (1 - 2\sqrt{2} + 2) = 12 - 6\sqrt{2} = 6(2 - \sqrt{2})$$ ### Pattern Recognition Always validate absolute root values against their domain restrictions to avoid including phantom solutions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations

### Related Formula Newton's Theorem for roots: If alpha, beta$\alpha, \beta$ satisfy ax^2+bx+c=0$ax^2+bx+c=0$ and S_n = alpha^n + beta^n$S_n = \alpha^n + \beta^n$, then: aS_n + bS_n-1 + cS_n-2 = 0$$aS_n + bS_{n-1} + cS_{n-2} = 0$$ ### Core Logic Apply Newton's Theorem directly to the first equation x^2+sqrt3x-16=0$x^2+\sqrt{3}x-16=0$ [cite: 1247]: P_n + sqrt3P_n-1 - 16P_n-2 = 0$$P_n + \sqrt{3}P_{n-1} - 16P_{n-2} = 0$$ [cite: 1248] For n=25$n=25$ [cite: 1248]: P_25 + sqrt3P_24 - 16P_23 = 0 implies P_25 + sqrt3P_24 = 16P_23$$P_{25} + \sqrt{3}P_{24} - 16P_{23} = 0 \implies P_{25} + \sqrt{3}P_{24} = 16P_{23}$$ [cite: 1248] Dividing both sides by 2P_23$2P_{23}$ [cite: 1249]: fracP_25 + sqrt3P_242P_23 = frac16P_232P_23 = 8$$\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$$ [cite: 1249] ### Step 1: Evaluation of the second part Now look at the second relation for x^2+3x-1=0$x^2+3x-1=0$ [cite: 1252]: gamma^2 + 3gamma - 1 = 0 implies gamma^2 - 1 = -3gamma$$\gamma^2 + 3\gamma - 1 = 0 \implies \gamma^2 - 1 = -3\gamma$$ delta^2 + 3delta - 1 = 0 implies delta^2 - 1 = -3delta$$\delta^2 + 3\delta - 1 = 0 \implies \delta^2 - 1 = -3\delta$$ Expand Q_25 - Q_23$Q_{25} - Q_{23}$ explicitly [cite: 1253]: Q_25 - Q_23 = (gamma^25 + delta^25) - (gamma^23 + delta^23) = gamma^23(gamma^2 - 1) + delta^23(delta^2 - 1)$$Q_{25} - Q_{23} = (\gamma^{25} + \delta^{25}) - (\gamma^{23} + \delta^{23}) = \gamma^{23}(\gamma^2 - 1) + \delta^{23}(\delta^2 - 1)$$ [cite: 1253] Substitute the root property transitions [cite: 1254]: = gamma^23(-3gamma) + delta^23(-3delta) = -3(gamma^24 + delta^24) = -3Q_24$$= \gamma^{23}(-3\gamma) + \delta^{23}(-3\delta) = -3(\gamma^{24} + \delta^{24}) = -3Q_{24}$$ [cite: 1254, 1255] Dividing by Q_24$Q_{24}$ gives [cite: 1256]: fracQ_25 - Q_23Q_24 = -3$$\frac{Q_{25} - Q_{23}}{Q_{24}} = -3$$ [cite: 1256] ### Step 2: Total Calculation Add both components evaluated above [cite: 1256]: textTotal Expression Value = 8 + (-3) = 5$$\text{Total Expression Value} = 8 + (-3) = 5$$ [cite: 1256] ### Pattern Recognition Expressions containing high sequential indices (like 25, 24, 23$25, 24, 23$) indicate recurrent functional applications. Always substitute fundamental characteristic polynomials instead of computing powers directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations

### Related Formula For quadratic equations with integer coefficients to have integral roots, the discriminant D = b^2 - 4ac$D = b^2 - 4ac$ must be a perfect square. ### Core Logic Rewrite using perfect square completing methods: x^2 + 4x + 4 = n + 4 implies (x + 2)^2 = n + 4 implies x = -2 pm sqrtn + 4$$x^2 + 4x + 4 = n + 4 \implies (x + 2)^2 = n + 4 \implies x = -2 \pm \sqrt{n + 4}$$ For x$x$ to be an integer, n + 4$n + 4$ must be a perfect square. Given range constraint 20 le n le 100$20 \le n \le 100$: 24 le n + 4 le 104$$24 \le n + 4 \le 104$$ ### Step 1: Identify Perfect Squares in Range Find perfect squares between 24 and 104: 5^2 = 25$5^2 = 25$ 6^2 = 36$6^2 = 36$ 7^2 = 49$7^2 = 49$ 8^2 = 64$8^2 = 64$ 9^2 = 81$9^2 = 81$ 10^2 = 100$10^2 = 100$ This gives exactly 6 distinct valid perfect squares. ### Step 2: Conclusion Thus, there are exactly 6 distinct integer values for n$n$. ### Pattern Recognition Completing the square provides intuitive bounds quicker than running full discriminant inequalities. Match integer root sets directly to explicit numerical sequence counts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 10 Mathematics: Quadratic Equations Class 11 Mathematics: Complex Numbers and Quadratic Equations

### Related Formula The definition of modulus function handles sub-intervals via critical points: |x - a| = begincases x - a & textif x ge a \\ -(x - a) & textif x < a endcases$$|x - a| = \begin{cases} x - a & \text{if } x \ge a \\ -(x - a) & \text{if } x < a \end{cases}$$ ### Core Logic The critical points are x = 2$x = 2$ and x = 3$x = 3$. We check the three distinct structural intervals: **Case I: x < 2$x < 2$** x(-(x - 2)) + 3(-(x - 3)) + 1 = 0$$x(-(x - 2)) + 3(-(x - 3)) + 1 = 0$$ -x^2 + 2x - 3x + 9 + 1 = 0 implies x^2 + x - 10 = 0$$-x^2 + 2x - 3x + 9 + 1 = 0 \implies x^2 + x - 10 = 0$$ x = frac-1 pm sqrt1 + 402 = frac-1 pm sqrt412$$x = \frac{-1 \pm \sqrt{1 + 40}}{2} = \frac{-1 \pm \sqrt{41}}{2}$$ Checking domain constraint x < 2$x < 2$: frac-1 - sqrt412 approx frac-1 - 6.42 = -3.7 < 2$\frac{-1 - \sqrt{41}}{2} \approx \frac{-1 - 6.4}{2} = -3.7 < 2$ (Valid root) frac-1 + sqrt412 approx frac-1 + 6.42 = 2.7 not< 2$\frac{-1 + \sqrt{41}}{2} \approx \frac{-1 + 6.4}{2} = 2.7 \not< 2$ (Rejected) ### Step 1: Intermediate Interval Check **Case II: 2 le x < 3$2 \le x < 3$** x(x - 2) + 3(-(x - 3)) + 1 = 0$$x(x - 2) + 3(-(x - 3)) + 1 = 0$$ x^2 - 2x - 3x + 9 + 1 = 0 implies x^2 - 5x + 10 = 0$$x^2 - 2x - 3x + 9 + 1 = 0 \implies x^2 - 5x + 10 = 0$$ Discriminant check: D = (-5)^2 - 4(1)(10) = 25 - 40 = -15 < 0$D = (-5)^2 - 4(1)(10) = 25 - 40 = -15 < 0$. No real roots exist in this interval. ### Step 2: Upper Interval Check **Case III: x ge 3$x \ge 3$** x(x - 2) + 3(x - 3) + 1 = 0$$x(x - 2) + 3(x - 3) + 1 = 0$$ x^2 - 2x + 3x - 9 + 1 = 0 implies x^2 + x - 8 = 0$$x^2 - 2x + 3x - 9 + 1 = 0 \implies x^2 + x - 8 = 0$$ x = frac-1 pm sqrt1 + 322 = frac-1 pm sqrt332$$x = \frac{-1 \pm \sqrt{1 + 32}}{2} = \frac{-1 \pm \sqrt{33}}{2}$$ Checking domain constraint x ge 3$x \ge 3$: frac-1 + sqrt332 approx frac-1 + 5.742 = 2.37 < 3$\frac{-1 + \sqrt{33}}{2} \approx \frac{-1 + 5.74}{2} = 2.37 < 3$ (Rejected) frac-1 - sqrt332 < 0$\frac{-1 - \sqrt{33}}{2} < 0$ (Rejected) Thus, only 1 valid real root satisfies the conditional layout across all ranges. ### Pattern Recognition Always perform case-by-case boundaries checks on algebraic roots found inside absolute modulus problems to discard ghost solutions quickly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations

### Related Formula For equations with repeating polynomial expressions, applying a variable substitution like t = P(x)$t = P(x)$ reduces high-degree polynomials down to standard quadratics. ### Core Logic Expand the linear binomial product in the given equation equation: (x-4)(x-5) = x^2 - 9x + 20$$(x-4)(x-5) = x^2 - 9x + 20$$ Rewrite the full expression in terms of the common variable pattern x^2 - 9x$x^2 - 9x$: (x^2 - 9x + 11)^2 - (x^2 - 9x + 20) = 3$$(x^2 - 9x + 11)^2 - (x^2 - 9x + 20) = 3$$ Let t = x^2 - 9x$t = x^2 - 9x$. Substituting this into the equation gives: (t + 11)^2 - (t + 20) = 3$$(t + 11)^2 - (t + 20) = 3$$ t^2 + 22t + 121 - t - 20 - 3 = 0$$t^2 + 22t + 121 - t - 20 - 3 = 0$$ t^2 + 21t + 98 = 0$$t^2 + 21t + 98 = 0$$ ### Step 1: Solve the Polynomial for Variable t Factorize the quadratic expression: (t + 14)(t + 7) = 0 implies t = -14 text or t = -7$$(t + 14)(t + 7) = 0 \implies t = -14 \text{ or } t = -7$$ ### Step 2: Back-substitute and Isolate Rational Roots Case 1: x^2 - 9x = -7 implies x^2 - 9x + 7 = 0$x^2 - 9x = -7 \implies x^2 - 9x + 7 = 0$ Check the discriminant value: D = (-9)^2 - 4(1)(7) = 81 - 28 = 53$D = (-9)^2 - 4(1)(7) = 81 - 28 = 53$ (not a perfect square, so the roots are irrational). Case 2: x^2 - 9x = -14 implies x^2 - 9x + 14 = 0$x^2 - 9x = -14 \implies x^2 - 9x + 14 = 0$ Factorize the quadratic expression: (x - 7)(x - 2) = 0 implies x = 7 text or x = 2$$(x - 7)(x - 2) = 0 \implies x = 7 \text{ or } x = 2$$ Both values are rational numbers. ### Step 3: Calculate the Product of Rational Roots Multiply the true rational roots together: textProduct = 7 cdot 2 = 14$$\text{Product} = 7 \cdot 2 = 14$$ ### Pattern Recognition Always check the discriminant D = b^2 - 4ac$D = b^2 - 4ac$ to filter out irrational radical components whenever the problem specifically asks for the product of *rational* roots only. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations