Quadratic Equations

• Branch 1: $x^2 - 5x + 6 = 0 \implies (x-2)(x-3) = 0 \implies x = 2, 3$
• Branch 2: $x^2 - 5x + 6 = -1 \implies x^2 - 5x + 7 = 0$. Let us analyze its discriminant: $D = (-5)^2 - 4(1)(7) = 25 - 28 = -3 < 0$. This branch produces no real roots.

• Case 1: The equation is linear. Set the leading coefficient to zero: $m - 1 = 0 \implies m = 1$. Substitute $m = 1$ back into the equation: $-4x + 1 = 0 \implies x = 0.25$. This gives exactly one unique solution, so $m = 1$ is valid.
• Case 2: The equation is quadratic ($m \neq 1$) and has equal roots ($D = 0$): \[ D = [-2(m+1)]^2 - 4(m-1)m = 4(m^2 + 2m + 1) - 4(m^2 - m) = 4(3m + 1) \] Set $D = 0 \implies 4(3m + 1) = 0 \implies m = -\frac{1}{3}$. Since $-\frac{1}{3} \neq 1$, this value is also completely valid.

Apply the standard quadratic formula directly, noting that $a = 1$, $b = -2i$, and $c = -3$: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-2i) \pm \sqrt{(-2i)^2 - 4(1)(-3)}}{2(1)} \] Calculate the values inside the radical sign: $(-2i)^2 = 4i^2 = -4$. Thus: \[ x = \frac{2i \pm \sqrt{-4 + 12}}{2} = \frac{2i \pm \sqrt{8}}{2} = \frac{2i \pm 2\sqrt{2}}{2} = i \pm \sqrt{2} \] Notice that because the coefficients are complex, the roots ($ \sqrt{2} + i $ and $-\sqrt{2} + i$) are not complex conjugates of each other. Final Answer: $x = \sqrt{2} + i, -\sqrt{2} + i$.

For the roots to be rational when the coefficients are integers, the discriminant $D = b^2 - 4ac$ must be a perfect square of an integer. Let $D = k^2$ for some integer $k$.
Since $a, b, c$ are odd integers, let us look at the expression modulo 8: Any odd integer squared is always congruent to 1 modulo 8 ($1^2 \equiv 1, 3^2 \equiv 9 \equiv 1, 5^2 \equiv 25 \equiv 1, 7^2 \equiv 49 \equiv 1$). Thus, $b^2 \equiv 1 \pmod 8$.
Now let us look at the second term, $4ac$: since both $a$ and $c$ are odd, their product $ac$ is also odd. Multiplying an odd number by 4 always yields: \[ 4ac = 4(\text{odd}) \equiv 4 \pmod 8 \] Now evaluate the full discriminant modulo 8: \[ D = b^2 - 4ac \equiv 1 - 4 \equiv -3 \equiv 5 \pmod 8 \] However, a perfect square integer $k^2$ can only be congruent to $0, 1,$ or $4$ modulo 8. It can never be congruent to $5 \pmod 8$. This is a contradiction, which proves that $D$ cannot be a perfect square. Therefore, the roots cannot be rational. Final Answer: Proved.

For the equation to have real roots, its discriminant must be non-negative ($D \ge 0$), and we must ensure the leading coefficient does not vanish ($p+4 \neq 0 \implies p \neq -4$). \[ D = (-2p)^2 - 4(p+4)(p-3) \ge 0 \] Expand the terms: \[ 4p^2 - 4(p^2 + p - 12) \ge 0 \implies 4p^2 - 4p^2 - 4p + 48 \ge 0 \] Simplify the linear inequality: \[ -4p + 48 \ge 0 \implies 4p \le 48 \implies p \le 12 \] Combine this with the quadratic condition $p \neq -4$ to write the final set: \[ p \in (-\infty, -4) \cup (-4, 12] \] Final Answer: $p \in (-\infty, -4) \cup (-4, 12]$.

Notice that the sum of the coefficients equals zero: $(b-c) + (c-a) + (a-b) = 0$. This means that $x = 1$ is automatically a root of this equation.
The problem states that the roots are equal ($D=0$), which means both roots must be exactly 1. Using the product of roots formula ($\alpha\beta = \frac{\text{constant term}}{\text{leading coefficient}}$): \[ 1 \times 1 = \frac{a-b}{b-c} \implies b-c = a-b \] Rearranging the terms gives: \[ 2b = a + c \] This proves that the coefficients $a, b, c$ form an Arithmetic Progression (AP). Final Answer: Proved.

Notice that $a_n = \alpha^n - \beta^n$ matches the exact operational structure of Newton's Sums. The base coefficients of our quadratic equation are $a = 1$, $b = -6$, and $c = -2$.
Write out the Newton's Sums recurrence relation for $n = 10$: \[ 1a_{10} - 6a_{10-1} - 2a_{10-2} = 0 \implies a_{10} - 6a_9 - 2a_8 = 0 \] Isolate the terms that appear in our target expression ($a_{10} - 2a_8$): \[ a_{10} - 2a_8 = 6a_9 \] Now substitute this identity directly into the numerator of our expression $E$: \[ E = \frac{6a_9}{2a_9} \] Since $\alpha \neq \beta$, $a_9 \neq 0$, so we can cancel $a_9$ from both the top and bottom: \[ E = \frac{6}{2} = 3 \] Final Answer: $3$.

• $S_0 = \alpha^0 + \beta^0 = 1 + 1 = 2$
• $S_1 = \alpha^1 + \beta^1 = -\frac{b}{a} = -\frac{-1}{1} = 1$

• $S_2 = S_1 + S_0 = 1 + 2 = 3$
• $S_3 = S_2 + S_1 = 3 + 1 = 4$
• $S_4 = S_3 + S_2 = 4 + 3 = 7$
• $S_5 = S_4 + S_3 = 7 + 4 = 11$

Identify the base coefficients from the given equation: $a = 5, b = 3, c = -7$.
Apply the Newton's Sums recurrence relation formula by choosing $n = 12$: \[ aV_{12} + bV_{11} + cV_{10} = 0 \] Substitute the values of the coefficients: \[ 5V_{12} + 3V_{11} - 7V_{10} = 0 \] Isolate $5V_{12}$ to complete the structural connection statement: \[ 5V_{12} = 7V_{10} - 3V_{11} \] Final Answer: $5V_{12} + 3V_{11} - 7V_{10} = 0$.

1. Discriminant condition ($D \ge 0$): \[ D = (-6a)^2 - 4(1)(9a^2 - 2a + 2) \ge 0 \implies 36a^2 - 36a^2 + 8a - 8 \ge 0 \implies 8a \ge 8 \implies a \ge 1 \]
2. Endpoint functional metric ($1 \cdot f(3) > 0$): \[ f(3) = 3^2 - 6a(3) + 9a^2 - 2a + 2 > 0 \implies 9 - 18a + 9a^2 - 2a + 2 > 0 \] Combine terms into a standard quadratic inequality: \[ 9a^2 - 20a + 11 > 0 \implies (9a - 11)(a - 1) > 0 \] This inequality is satisfied for: $a \in (-\infty, 1) \cup (\frac{11}{9}, \infty)$.
3. Vertex position condition ($-\frac{b}{2a} > 3$): \[ -\frac{-6a}{2(1)} > 3 \implies 3a > 3 \implies a > 1 \]

The condition for exactly one root to lie inside an open interval $(k_1, k_2)$ is that the values of the function at the endpoints must have opposite signs: $f(k_1) \cdot f(k_2) < 0$.
Let $f(x) = x^2 - kx + 4$. Calculate the values at the endpoints $x = 1$ and $x = 3$: \[ f(1) = 1^2 - k(1) + 4 = 5 - k \] \[ f(3) = 3^2 - k(3) + 4 = 13 - 3k \] Set up the product inequality: \[ (5 - k)(13 - 3k) < 0 \] Multiply by $-1$ twice to change the signs of both linear factors, which leaves the direction of the inequality sign unchanged: \[ (k - 5)(3k - 13) < 0 \] Using the wavy-curve method, the product is negative between the two roots $k = \frac{13}{3}$ and $k = 5$. Thus, $k \in (\frac{13}{3}, 5)$. (Note: We must double-check that the endpoints do not create a condition where a root lands exactly on an endpoint, which is excluded by the open interval). Final Answer: $k \in (\frac{13}{3}, 5)$.

1. Discriminant condition ($D \ge 0$): \[ D = (-2m)^2 - 4(1)(m^2 - 1) = 4m^2 - 4m^2 + 4 = 4 > 0 \] Since $D = 4$, the roots are real and distinct for all values of $m$.
2. Endpoint values must be positive ($f(-2) > 0$ and $f(4) > 0$): \[ f(-2) = (-2)^2 - 2m(-2) + m^2 - 1 = m^2 + 4m + 3 > 0 \implies (m+1)(m+3) > 0 \] This gives the interval: $m \in (-\infty, -3) \cup (-1, \infty)$. \[ f(4) = 4^2 - 2m(4) + m^2 - 1 = m^2 - 8m + 15 > 0 \implies (m-3)(m-5) > 0 \] This gives the interval: $m \in (-\infty, 3) \cup (5, \infty)$. Intersecting these two intervals yields: $m \in (-\infty, -3) \cup (-1, 3) \cup (5, \infty)$.
3. Vertex position condition ($-2 < -\frac{b}{2a} < 4$): \[ -2 < -\frac{-2m}{2(1)} < 4 \implies -2 < m < 4 \]

First, verify that the denominator $x^2 + x + 1$ is never zero. Its discriminant is $1^2 - 4(1)(1) = -3 < 0$, so the denominator is strictly positive for all real $x$. The function is well-defined everywhere.
Cross-multiply to rearrange the expression: \[ y(x^2 + x + 1) = x^2 - x + 1 \implies yx^2 + yx + y = x^2 - x + 1 \] Group the terms into a standard quadratic equation in terms of $x$: \[ (y - 1)x^2 + (y + 1)x + (y - 1) = 0 \] Since $x$ is a real number ($x \in \mathbb{R}$), the discriminant of this quadratic equation must be non-negative ($D \ge 0$): \[ D = (y + 1)^2 - 4(y - 1)(y - 1) \ge 0 \] Expand the terms: \[ (y^2 + 2y + 1) - 4(y^2 - 2y + 1) \ge 0 \implies y^2 + 2y + 1 - 4y^2 + 8y - 4 \ge 0 \] Combine like terms to simplify the inequality: \[ -3y^2 + 10y - 3 \ge 0 \] Multiply by $-1$ and flip the direction of the inequality sign: \[ 3y^2 - 10y + 3 \le 0 \implies (3y - 1)(y - 3) \le 0 \] Using the wavy-curve method, the inequality is satisfied for values of $y$ between the two roots: \[ y \in \left[ \frac{1}{3}, 3 \right] \] We must check the boundary value where the leading coefficient vanishes ($y - 1 = 0 \implies y = 1$). If $y = 1$, the equation becomes $2x = 0 \implies x = 0$, which is a valid real number. Therefore, $y = 1$ is included in the range. Final Answer: Range $\in [\frac{1}{3}, 3]$.

Set the function equal to $y$: \[ y = \frac{x}{x^2 - 5x + 9} \implies y(x^2 - 5x + 9) = x \] Rearrange the terms into a standard quadratic equation in $x$: \[ yx^2 - (5y + 1)x + 9y = 0 \] Since $x \in \mathbb{R}$, enforce the condition $D \ge 0$: \[ D = [-(5y + 1)]^2 - 4(y)(9y) \ge 0 \] Expand and simplify the terms: \[ (25y^2 + 10y + 1) - 36y^2 \ge 0 \implies -11y^2 + 10y + 1 \ge 0 \] Multiply by $-1$ and flip the inequality sign: \[ 11y^2 - 10y - 1 \le 0 \implies (11y + 1)(y - 1) \le 0 \] Using the wavy-curve method, the interval is: \[ y \in \left[ -\frac{1}{11}, 1 \right] \] Thus, the minimum value is $-\frac{1}{11}$ and the maximum value is $1$. Final Answer: Minimum = $-\frac{1}{11}$, Maximum = $1$.

Cross-multiply and rearrange into a quadratic equation in $x$: \[ (y - 1)x^2 + (4y - 2)x + (3cy - c) = 0 \] Enforce the condition $D \ge 0$ for real $x$: \[ D = (4y - 2)^2 - 4(y - 1)c(3y - 1) \ge 0 \] Divide the inequality by 4: \[ (2y - 1)^2 - c(y - 1)(3y - 1) \ge 0 \] Expand the terms to form a quadratic inequality in terms of $y$: \[ (4y^2 - 4y + 1) - c(3y^2 - 4y + 1) \ge 0 \implies (4 - 3c)y^2 - 4(1 - c)y + (1 - c) \ge 0 \] The problem states that this inequality must hold true for all real values of $y$ ($y \in \mathbb{R}$). For a quadratic expression to be non-negative for all real values, its leading coefficient must be positive ($4 - 3c > 0 \implies c < \frac{4}{3}$) and its discriminant relative to $y$ must be less than or equal to zero ($D_y \le 0$): \[ D_y = [-4(1-c)]^2 - 4(4-3c)(1-c) \le 0 \] Divide by 4 and factor out $(1-c)$: \[ 4(1-c)^2 - (4-3c)(1-c) \le 0 \implies (1-c)[4(1-c) - (4-3c)] \le 0 \] Simplify the inner expression: \[ (1-c)[4 - 4c - 4 + 3c] \le 0 \implies (1-c)(-c) \le 0 \implies c(c-1) \le 0 \] This inequality is satisfied for the interval $c \in [0, 1]$. Since this entire interval is strictly less than $\frac{4}{3}$, it is completely valid. Final Answer: $c \in [0, 1]$.

• First change: from $+1$ to $-1$ (1 sign change).
• Second change: from $-1$ to $+1$ (1 sign change).

Let $P(x) = x^3 - 3x + 1$. Since $P(x)$ is a polynomial function, it is continuous everywhere. Let us evaluate the value of the function at the interval endpoints $x = 0$ and $x = 1$: \[ P(0) = 0^3 - 3(0) + 1 = 1 \quad (\text{which is } > 0) \] \[ P(1) = 1^3 - 3(1) + 1 = -1 \quad (\text{which is } < 0) \] Calculate the product of the endpoint values: \[ P(0) \cdot P(1) = 1 \times (-1) = -1 < 0 \] Since the function values at the endpoints have opposite signs, the Intermediate Value Theorem (IVT) guarantees that the continuous curve must cross the x-axis at least once inside the interval. Therefore, there is at least one real root in $(0, 1)$. Final Answer: Proved.

• Branch 1: $x = 0$. This is a valid real root.
• Branch 2: $x^6 + x^4 + x^2 + 1 = 0$.

Let $\alpha$ be the shared common root. Substitute $\alpha$ into both equations: \[ \alpha^2 - k\alpha - 21 = 0 \quad \text{--- (Equation 1)} \] \[ \alpha^2 - 3k\alpha + 35 = 0 \quad \text{--- (Equation 2)} \] Subtract Equation 1 from Equation 2 to eliminate the quadratic $\alpha^2$ term: \[ (\alpha^2 - 3k\alpha + 35) - (\alpha^2 - k\alpha - 21) = 0 \implies -2k\alpha + 56 = 0 \] Isolate the common root $\alpha$ in terms of $k$: \[ 2k\alpha = 56 \implies \alpha = \frac{28}{k} \] Now substitute this expression for $\alpha$ back into Equation 1: \[ \left(\frac{28}{k}\right)^2 - k\left(\frac{28}{k}\right) - 21 = 0 \implies \frac{784}{k^2} - 28 - 21 = 0 \] Simplify the equation: \[ \frac{784}{k^2} = 49 \implies k^2 = \frac{784}{49} = 16 \implies k = \pm 4 \] Both values of $k$ are valid and create exactly one common root. Final Answer: $k = \pm 4$.

Let us analyze the nature of the roots of the first equation, $x^2 + 2x + 3 = 0$: Its discriminant is $D = 2^2 - 4(1)(3) = 4 - 12 = -8 < 0$. This means the roots are non-real complex conjugates. Since the problem states that the second equation has real coefficients ($a,b,c \in \mathbb{R}$), its imaginary roots must also emerge as conjugate pairs.
Therefore, if the two equations share one imaginary root, they must share the other root as well. This means both roots are common, so the equations are identical. Equate the ratios of their coefficients: \[ \frac{a}{1} = \frac{b}{2} = \frac{c}{3} \implies a:b:c = 1:2:3 \] This is a standard trap question on both the JEE Main and Advanced exams. Final Answer: $1:2:3$.

Let $\alpha$ be the shared common root. Substitute $\alpha$ into both equations: \[ \alpha^2 + a\alpha + b = 0 \] \[ \alpha^2 + b\alpha + a = 0 \] Subtract the two equations to eliminate $\alpha^2$: \[ (a - b)\alpha + (b - a) = 0 \implies (a - b)\alpha = a - b \] Since the problem states $a \neq b$, we can divide both sides by $(a - b)$ to find the common root: \[ \alpha = 1 \] Now substitute the value $\alpha = 1$ back into the first equation: \[ 1^2 + a(1) + b = 0 \implies 1 + a + b = 0 \] This gives the required condition: $a + b + 1 = 0$. Final Answer: $a + b + 1 = 0$.

Identify the coefficients of the quadratic function: $a = 3$, $b = -4$, and $c = 7$. Since the leading coefficient is $a = 3 > 0$, the parabola opens upward, meaning the function has a global minimum value at its vertex.
Let us first calculate the discriminant $D$: \[ D = b^2 - 4ac = (-4)^2 - 4(3)(7) = 16 - 84 = -68 \] Now evaluate the minimum value using the formula $-\frac{D}{4a}$: \[ \text{Minimum Value} = -\frac{-68}{4(3)} = \frac{68}{12} = \frac{17}{3} \] This minimum value occurs at the vertex x-coordinate: \[ x = -\frac{b}{2a} = -\frac{-4}{2(3)} = \frac{4}{6} = \frac{2}{3} \] Final Answer: Minimum value is $\frac{17}{3}$, which occurs at $x = \frac{2}{3}$.

Rearrange the expression into standard form: $y = -4x^2 + 8x + 5$. Identify the coefficients: $a = -4$, $b = 8$, and $c = 5$.
Since the leading coefficient is $a = -4 < 0$, the parabola opens downward, meaning the function has a global maximum value at its vertex. Calculate the discriminant $D$: \[ D = b^2 - 4ac = 8^2 - 4(-4)(5) = 64 + 80 = 144 \] Evaluate the maximum value using the vertex formula $-\frac{D}{4a}$: \[ \text{Maximum Value} = -\frac{144}{4(-4)} = \frac{144}{16} = 9 \] This peak value occurs at the x-coordinate: \[ x = -\frac{b}{2a} = -\frac{8}{2(-4)} = \frac{-8}{-8} = 1 \] Final Answer: Maximum value is $9$, which occurs at $x = 1$.

A common mistake is to only calculate the values at the endpoints of the interval. We must also check if the vertex falls inside the restricted domain interval. Identify the vertex x-coordinate: \[ x_v = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1 \] Since $1 \in [0, 3]$, the vertex lies inside the restricted domain. Since $a = 1 > 0$, the function achieves its global minimum value at this vertex: \[ \text{Minimum Value} = f(1) = 1^2 - 2(1) + 5 = 4 \] Now evaluate the value of the function at the domain endpoints $x = 0$ and $x = 3$ to find the maximum value: \[ f(0) = 0^2 - 2(0) + 5 = 5 \] \[ f(3) = 3^2 - 2(3) + 5 = 9 - 6 + 5 = 8 \] Comparing the endpoint values ($5$ and $8$), the maximum value of the function on this interval is $8$.
Therefore, the range of the function over the restricted domain is $[4, 8]$. Final Answer: Range $\in [4, 8]$.