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

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

Core Logic

For the denominator x² - 8x + 32$x^2 - 8x + 32$, D = 64 - 128 < 0$D = 64 - 128 < 0$ and a = 1 > 0$a = 1 > 0$. Thus, x² - 8x + 32 > 0 x in R$x^2 - 8x + 32 > 0 \forall x \in \mathbb{R}$.

Step 1: Constraint on Numerator

Since the denominator is always positive, the numerator must be strictly negative for all x in R$x \in \mathbb{R}$.

This requires a < 0$a < 0$ and D < 0$D < 0$.

Step 2: Conclusion

Since a$a$ must be strictly less than 0$0$, there are no positive integral values of a$a$ that satisfy the condition. Hence, S$S$ is an empty set. Number of elements is 0.

Chapter Mix

Class 11 Maths: Quadratic Equations