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

Core Logic

Restructure the equation to form a quadratic in |√(x) - 3|$|\sqrt{x} - 3|$. Notice that (x - 6√(x) + 9) = (√(x) - 3)² = |√(x) - 3|²$(x - 6\sqrt{x} + 9) = (\sqrt{x} - 3)^2 = |\sqrt{x} - 3|^2$. Rewrite the given equation:

Step 1: Solve the Quadratic

Let u = |√(x) - 3|$u = |\sqrt{x} - 3|$. The equation is u² - (2 - √(3))u - 2√(3) = 0$u^2 - (2 - \sqrt{3})u - 2\sqrt{3} = 0$. Factorizing gives:

So, u = 2$u = 2$ or u = -√(3)$u = -\sqrt{3}$. Since u = |√(x) - 3|$u = |\sqrt{x} - 3|$ cannot be negative, we reject u = -√(3)$u = -\sqrt{3}$. Thus, |√(x) - 3| = 2$|\sqrt{x} - 3| = 2$.

Step 2: Find x (Roots)

Solve |√(x) - 3| = 2$|\sqrt{x} - 3| = 2$:

Squaring gives x = 25$x = 25$ or x = 1$x = 1$. Given α < β$\alpha < \beta$, we have α = 1$\alpha = 1$ and β = 25$\beta = 25$.

Step 3: Evaluate Target Expression

Now compute √((β)/(α)) + √(αβ)$\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$:

Pattern Recognition

Grouping algebraic terms (like x - 6√(x)$x - 6\sqrt{x}$) and a lone constant (+9$+9$) to form perfect squares is a hallmark of radical equations disguised as quadratics.

Chapter Mix

Class 11 Maths: Quadratic Equations

Related Formula

Core Logic

Factor the arguments in the logarithmic equation: 6x² + 28x + 30 = (x+3)(6x+10)$6x^2 + 28x + 30 = (x+3)(6x+10)$ x² + 6x + 9 = (x+3)²$x^2 + 6x + 9 = (x+3)^2$

Substitute these into the equation:

Step 1: Variable Substitution

Let A = (x+3)(6x+10)$A = \log_{(x+3)}(6x+10)$. The equation transforms to:

Substitute A = 2$A = 2$ back:

Step 2: Checking Domain Validity

For x = 1$x = 1$: Base x+3 = 4 > 0, ≠ 1$x+3 = 4 > 0, \neq 1$. Base 6x+10 = 16 > 0, ≠ 1$6x+10 = 16 > 0, \neq 1$. Valid solution.

For x = -1$x = -1$: Base x+3 = 2 > 0, ≠ 1$x+3 = 2 > 0, \neq 1$. Base 6x+10 = 4 > 0, ≠ 1$6x+10 = 4 > 0, \neq 1$. Valid solution.

Sum of all real roots = 1 + (-1) = 0$= 1 + (-1) = 0$.

Pattern Recognition

When dealing with logarithms containing polynomial bases and arguments, always check if they are directly factorable into each other. A substitution like A + B/A = C$A + B/A = C$ will often emerge.

Chapter Mix

Class 11 Maths: Quadratic Equations Class 11 Maths: Functions

Related Formula

Core Logic

Step 1: Case 1 (x > 1)

For x > 1$x > 1$: x(x+3) + (x-1) - 2 = 0$x(x+3) + (x-1) - 2 = 0$ x² + 3x + x - 3 = 0 ⇒ x² + 4x - 3 = 0$x^2 + 3x + x - 3 = 0 \Rightarrow x^2 + 4x - 3 = 0$ x = -4 ± √(16 + 12)2 = -2 ± √(7)$x = \frac{-4 \pm \sqrt{16 + 12}}{2} = -2 \pm \sqrt{7}$ Since √(7) ≈ 2.64$\sqrt{7} \approx 2.64$, x = -2 + 2.64 = 0.64$x = -2 + 2.64 = 0.64$, which is not > 1$> 1$. Both rejected.

Step 2: Case 2 (-3 <= x <= 1)

For -3 ≤ x ≤ 1$-3 \le x \le 1$: x(x+3) - (x-1) - 2 = 0$x(x+3) - (x-1) - 2 = 0$ x² + 3x - x + 1 - 2 = 0 ⇒ x² + 2x - 1 = 0$x^2 + 3x - x + 1 - 2 = 0 \Rightarrow x^2 + 2x - 1 = 0$ x = -2 ± √(4 + 4)2 = -1 ± √(2)$x = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2}$ √(2) ≈ 1.41$\sqrt{2} \approx 1.41$. x = -1 + 1.41 = 0.41$x = -1 + 1.41 = 0.41$ (Accepted) x = -1 - 1.41 = -2.41$x = -1 - 1.41 = -2.41$ (Accepted) (2 solutions)

Step 3: Case 3 (x < -3)

For x < -3$x < -3$: x(-x-3) - (x-1) - 2 = 0$x(-x-3) - (x-1) - 2 = 0$ -x² - 3x - x + 1 - 2 = 0 ⇒ -x² - 4x - 1 = 0 ⇒ x² + 4x + 1 = 0$-x^2 - 3x - x + 1 - 2 = 0 \Rightarrow -x^2 - 4x - 1 = 0 \Rightarrow x^2 + 4x + 1 = 0$ x = -4 ± √(16 - 4)2 = -2 ± √(3)$x = \frac{-4 \pm \sqrt{16 - 4}}{2} = -2 \pm \sqrt{3}$ x = -2 + 1.732 = -0.268$x = -2 + 1.732 = -0.268$ (Rejected, not < -3$< -3$) x = -2 - 1.732 = -3.732$x = -2 - 1.732 = -3.732$ (Accepted, < -3$< -3$) (1 solution) Total valid real solutions = 2 + 1 = 3$2 + 1 = 3$.

Pattern Recognition

Modulus equations involving polynomials are best resolved by strictly zoning the number line via critical points, verifying root validity against the respective zone boundaries.

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations

Related Formula

Core Logic

Substitute x² = t$x^2 = t$. The equation becomes a quadratic in t$t$:

For the original quartic equation x⁴ - ax² + 9 = 0$x^4 - ax^2 + 9 = 0$ to have 4 real and distinct roots, the quadratic equation in t$t$ must have 2 distinct positive real roots (since x = ± √(t)$x = \pm \sqrt{t}$ requires t > 0$t > 0$).

Step 1: Discriminant Condition

Condition 1: Roots must be real and distinct (D > 0$D > 0$)

a² - 36 > 0$a^2 - 36 > 0$

Step 2: Location of Roots Condition

Condition 2: Sum of roots must be positive (since both roots are positive)

Condition 3: Product of roots must be positive

Step 3: Intersection and Conclusion

Taking the intersection of all conditions: a in (-∞, -6) (6, ∞)$a \in (-\infty, -6) \cup (6, \infty)$ AND a > 0$a > 0$.

Intersection yields: a in (6, ∞)$a \in (6, \infty)$.

The smallest positive integral value in this interval is 7$7$.

Pattern Recognition

Bi-quadratic equations x⁴ + Bx² + C = 0$x^4 + Bx^2 + C = 0$ map cleanly to t² + Bt + C = 0$t^2 + Bt + C = 0$. The nature of x$x$ roots depends entirely on the signs of t$t$ roots. 4 real distinct x$x$ roots ≡$\equiv$ 2 positive distinct t$t$ roots.

Chapter Mix

Class 11 Maths: Quadratic Equations

Related Formula

For a quadratic equation ax² + bx + c = 0$ax^2 + bx + c = 0$ to have equal roots, its discriminant must be zero:

Perpendicular distance of point (x₀, y₀)$(x_0, y_0)$ from line Ax + By + C = 0$Ax + By + C = 0$ is:

Core Logic

Let's expand the given equation:

Let λ = 12-k$\lambda = 12-k$. The quadratic equation is:

Step 1: Finding k$k$

Set the discriminant to zero:

Since λ ≠ 0$\lambda \neq 0$ (otherwise it is not quadratic and has no roots): λ = -2$\lambda = -2$

Thus:

Step 2: Calculating perpendicular distance

The point of interest is:

Distance from the line 3x + 4y + 5 = 0$3x + 4y + 5 = 0$:

Pattern Recognition

In quadratic equation analysis, substitution of variable coefficients with parameter λ$\lambda$ keeps calculations clean and helps identify constraints such as λ ≠ 0$\lambda \neq 0$ at early stages.

Chapter Mix

Class 11 Mathematics: Quadratic Equations Class 10 Mathematics: Coordinate Geometry