If the set of all a in R$a \in \mathbb{R}$, for which the equation 2x² + (a - 5)x + 15 = 3a$2x^2 + (a - 5)x + 15 = 3a$ has no real root, is the interval (α, β)$(\alpha, \beta)$, and X = x in Z : α < x < β$X = \{x \in \mathbb{Z} : \alpha < x < \beta\}$, then Σx in X x²$\sum_{x \in X} x^2$ is equal to

Related Formula

For quadratic equations with integer coefficients to have integral roots, the discriminant D = b² - 4ac$D = b^2 - 4ac$ must be a perfect square.

Core Logic

Rewrite using perfect square completing methods:

For x$x$ to be an integer, n + 4$n + 4$ must be a perfect square. Given range constraint 20 ≤ n ≤ 100$20 \le n \le 100$:

Step 1: Identify Perfect Squares in Range

Find perfect squares between 24 and 104: 5² = 25$5^2 = 25$ 6² = 36$6^2 = 36$ 7² = 49$7^2 = 49$ 8² = 64$8^2 = 64$ 9² = 81$9^2 = 81$ 10² = 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.

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:

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$x < 2$

Checking domain constraint x < 2$x < 2$: -1 - √(41)2 ≈ (-1 - 6.4)/(2) = -3.7 < 2$\frac{-1 - \sqrt{41}}{2} \approx \frac{-1 - 6.4}{2} = -3.7 < 2$ (Valid root) -1 + √(41)2 ≈ (-1 + 6.4)/(2) = 2.7 < 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 ≤ x < 3$2 \le x < 3$

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 ≥ 3$x \ge 3$

Checking domain constraint x ≥ 3$x \ge 3$: -1 + √(33)2 ≈ (-1 + 5.74)/(2) = 2.37 < 3$\frac{-1 + \sqrt{33}}{2} \approx \frac{-1 + 5.74}{2} = 2.37 < 3$ (Rejected) -1 - √(33)2 < 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.

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:

Rewrite the full expression in terms of the common variable pattern x² - 9x$x^2 - 9x$:

Let t = x² - 9x$t = x^2 - 9x$. Substituting this into the equation gives:

Step 1: Solve the Polynomial for Variable t

Factorize the quadratic expression:

Step 2: Back-substitute and Isolate Rational Roots

Case 1: x² - 9x = -7 x² - 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: x² - 9x = -14 x² - 9x + 14 = 0$x^2 - 9x = -14 \implies x^2 - 9x + 14 = 0$ Factorize the quadratic expression:

Both values are rational numbers.

Step 3: Calculate the Product of Rational Roots

Multiply the true rational roots together:

Pattern Recognition

Always check the discriminant D = b² - 4ac$D = b^2 - 4ac$ to filter out irrational radical components whenever the problem specifically asks for the product of rational roots only.

Chapter Mix

Class 11 Mathematics: Quadratic Equations

Related Formula

Core Logic

Let 1√(x) = α$\frac{1}{\sqrt{x}} = \alpha$. The equation reduces to a product of two quadratics:

Step 1: Factorize the components

First quadratic: 9α² - 9α + 2 = 0 (3α - 2)(3α - 1) = 0 α = (2)/(3), (1)/(3)$9\alpha^2 - 9\alpha + 2 = 0 \implies (3\alpha - 2)(3\alpha - 1) = 0 \implies \alpha = \frac{2}{3}, \frac{1}{3}$ Second quadratic: 2α² - 7α + 3 = 0 (2α - 1)(α - 3) = 0 α = (1)/(2), 3$2\alpha^2 - 7\alpha + 3 = 0 \implies (2\alpha - 1)(\alpha - 3) = 0 \implies \alpha = \frac{1}{2}, 3$

Step 2: Solve for x

Since α = 1√(x) x = (1)/(α²)$\alpha = \frac{1}{\sqrt{x}} \implies x = \frac{1}{\alpha^2}$. For α = (1)/(3) x = 9$\alpha = \frac{1}{3} \implies x = 9$ For α = (1)/(2) x = 4$\alpha = \frac{1}{2} \implies x = 4$ For α = (2)/(3) x = (9)/(4)$\alpha = \frac{2}{3} \implies x = \frac{9}{4}$ For α = 3 x = (1)/(9)$\alpha = 3 \implies x = \frac{1}{9}$ All 4 values are positive and valid.

Pattern Recognition

Always check constraints first (x > 0$x > 0$ due to √(x)$\sqrt{x}$ in denominator). Since all roots α > 0$\alpha > 0$, every single algebraic root maps to a real distinct solution.

Chapter Mix

Class 11 Mathematics: Quadratic Equations

Related Formula

Conjugate Surd Identity:

Core Logic

Observe the base components of the exponents:

Therefore, we can express one base as the reciprocal of the other:

The equation becomes:

Step 1: Formulate the Quadratic Equation

Let (√(3) + √(2))x = t$(\sqrt{3} + \sqrt{2})^{x} = t$. Then:

Solving for t$t$ using the quadratic formula:

Step 2: Solve for x

Notice that (5 ± 2√(6))$(5 \pm 2\sqrt{6})$ can be written as square powers of the original base:

Thus, we have:

• For t = 5 + 2√(6) (√(3) + √(2))x = (√(3) + √(2))² x = 2$t = 5 + 2\sqrt{6} \implies (\sqrt{3} + \sqrt{2})^{x} = (\sqrt{3} + \sqrt{2})^{2} \implies x = 2$
• For t = 5 - 2√(6) (√(3) + √(2))x = (√(3) - √(2))² = (√(3) + √(2))⁻² x = -2$t = 5 - 2\sqrt{6} \implies (\sqrt{3} + \sqrt{2})^{x} = (\sqrt{3} - \sqrt{2})^{2} = (\sqrt{3} + \sqrt{2})^{-2} \implies x = -2$

Therefore, the distinct real solutions are x = 2$x = 2$ and x = -2$x = -2$. The number of elements in set S$S$ is 2$2$.

Pattern Recognition

Sees: Rational conjugate bases added with inverse matching variables. Shortcut: Whenever you see an equation of the form A^x + B^x = C$A^x + B^x = C$ where AB = 1$AB = 1$, the solution will always be symmetric (± x₀$\pm x_0$). Checking x=2$x=2$ explicitly gives (√(3)+√(2))² + (√(3)-√(2))² = (5+2√(6)) + (5-2√(6)) = 10$(\sqrt{3}+\sqrt{2})^2 + (\sqrt{3}-\sqrt{2})^2 = (5+2\sqrt{6}) + (5-2\sqrt{6}) = 10$, confirming ± 2$\pm 2$ immediately.

Chapter Mix

Class 11 Mathematics: Quadratic Equations Class 9 Mathematics: Number Systems (Rationalization)