Let S = x³ + ax² + bx + c : a, b, c in N and a, b, c ≤ 20$S = \{ x^3 + ax^2 + bx + c : a, b, c \in \mathbb{N} \text{ and } a, b, c \leq 20 \}$ be a set of polynomials. Then the number of polynomials in S$S$, which are divisible by x² + 2$x^2 + 2$, is

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)

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Core Logic

Given equation:

This factors into two possibilities:

• x = 0$x = 0$
• x² + 3|x| + 5|x - 1| + 6|x - 2| = 0$x^2 + 3|x| + 5|x - 1| + 6|x - 2| = 0$

Step 1: Evaluating the Modulus Term

Look at the second factor: f(x) = x² + 3|x| + 5|x - 1| + 6|x - 2|$f(x) = x^2 + 3|x| + 5|x - 1| + 6|x - 2|$. Notice that all terms inside are strictly non-negative:

• x² ≥ 0$x^2 \ge 0$
• 3|x| ≥ 0$3|x| \ge 0$
• 5|x - 1| ≥ 0$5|x - 1| \ge 0$
• 6|x - 2| ≥ 0$6|x - 2| \ge 0$

For the sum to be 0$0$, every single term must be simultaneously zero. x² = 0 x = 0$x^2 = 0 \implies x = 0$ However, if x = 0$x = 0$, then 5|x-1| = 5(1) = 5 ≠ 0$5|x-1| = 5(1) = 5 \neq 0$. Therefore, there is no real value of x$x$ that makes this entire second factor equal to zero.

Step 2: Conclusion

The only valid solution to the equation is x = 0$x = 0$ from the first factor. Thus, there is exactly 1$1$ real solution.

Pattern Recognition

A sum of absolute values and squares set to 0 requires all individual components to hit 0 concurrently. If they have different zero-nodes (0, 1, 2$0, 1, 2$), the sum can never be 0$0$.

Chapter Mix

Class 11 Maths: Quadratic Equations

Core Logic

Given equation: (a²+b²)x² - 2b(a+c)x + b²+c² = 0$(a^2+b^2)x^2 - 2b(a+c)x + b^2+c^2 = 0$. Expand and rearrange into perfect squares:

Since squares must be non-negative, each term is zero:

Thus, b = ax$b = ax$ and c = bx = ax²$c = bx = ax^2$. Since a,b,c$a,b,c$ form a triangle, the triangle inequality holds:

• a + b > c a + ax > ax² x² - x - 1 < 0 1-√(5)2 < x < 1+√(5)2$a + b > c \implies a + ax > ax^2 \implies x^2 - x - 1 < 0 \implies \frac{1-\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2}$
• a + c > b a + ax² > ax x² - x + 1 > 0$a + c > b \implies a + ax^2 > ax \implies x^2 - x + 1 > 0$ (Always true for real x$x$)
• b + c > a ax + ax² > a x² + x - 1 > 0 x > -1+√(5)2$b + c > a \implies ax + ax^2 > a \implies x^2 + x - 1 > 0 \implies x > \frac{-1+\sqrt{5}}{2}$ or x < -1-√(5)2$x < \frac{-1-\sqrt{5}}{2}$.

Taking the intersection (and noting x > 0$x > 0$ since sides are positive):

So, α = √(5)-12$\alpha = \frac{\sqrt{5}-1}{2}$ and β = √(5)+12$\beta = \frac{\sqrt{5}+1}{2}$. Calculate 12(α² + β²)$12(\alpha^2 + \beta^2)$:

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines

Related Formula

Core Logic

Given f(x) = (a + b - 2c)x² + (b + c - 2a)x + (c + a - 2b) = 0$f(x) = (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0$. Substituting x = 1$x = 1$:

Thus, one root is 1$1$. Let the other root be α$\alpha$.

Step 1: Find the other root

Product of roots = (c + a - 2b)/(a + b - 2c)$\frac{c + a - 2b}{a + b - 2c}$. Since one root is 1$1$, we have:

Step 2: Analyze Statement (I)

If -1 < α < 0$-1 < \alpha < 0$:

This implies b > (a + c)/(2)$b > \frac{a + c}{2}$ and b + c < 2a$b + c < 2a$. Therefore, b$b$ cannot be the Geometric Mean of a$a$ and c$c$. Statement (I) is true.

Step 3: Analyze Statement (II)

If 0 < α < 1$0 < \alpha < 1$:

This gives b > c$b > c$ and b < (a + c)/(2)$b < \frac{a + c}{2}$. Therefore, b$b$ may be the Geometric Mean between a$a$ and c$c$. Statement (II) is true.

Pattern Recognition

When coefficients in a quadratic equation are cyclic and sum to 0, one root is always 1$1$. The other root is directly c/a$c/a$.

Chapter Mix

Class 11 Maths: Quadratic Equations Class 11 Maths: Sequences and Series