Quadratic Equation And Inequalities – JEE Main MathematicsPractice Questions & PYQs

The sum of the coefficients of the quadratic equation a(b-c)x^2 + b(c-a)x + c(a-b) = 0 is 0, which implies that x=1 is always a root. Given the condition that the equation has equal roots, both roots must be 1, meaning the product of the roots c(a-b)/a(b-c) must also be 1. This condition leads to the equality 2ac = b(a+c). Substituting the given values a+c=15 and b=36/5 into this equation yields 2ac = 108, or ac = 54. Using the algebraic identity a^{2+}c^2 = (a+c)^2 - 2ac, we find 15^2 - 2(54) = 225 - 108 = 117.

By substituting $t = \frac{1}{\sqrt{x}}$, the given expression transforms into a pair of quadratic equations in $t$: $(9t^2 - 9t + 2)(2t^2 - 7t + 3) = 0$. Factoring these yields $t = \frac{2}{3}, \frac{1}{3}$ and $t = \frac{1}{2}, 3$ respectively. Since $\sqrt{x} = \frac{1}{t}$, the corresponding values for $\sqrt{x}$ are $\frac{3}{2}, 3, 2, \frac{1}{3}$. Squaring these gives $x = \frac{9}{4}, 9, 4, \frac{1}{9}$, resulting in four distinct real solutions.

For a quadratic equation to have positive roots, the discriminant must be non-negative, the product of the roots must be positive, and the sum of the roots must be positive. Evaluating the discriminant $D = 4(a-3)^2 - 4(9)(1-a) \ge 0$ gives $a \in (-\infty , -3] \cup [0, \infty )$, the root product condition $\frac{9}{1-a} > 0$ requires $a < 1$, and the sum condition $\frac{-2(a-3)}{1-a} > 0$ restricts $a$ to $(-\infty , 1) \cup (3, \infty )$. The intersection of these intervals is $a \in (-\infty , -3] \cup [0, 1)$, identifying $\alpha=3$, $\beta=0$, and $\gamma=1$, which makes $2\alpha+\beta+\gamma = 7$.

The condition for a quadratic equation to have equal roots is that its discriminant must be zero. Expressing the given equation as x^{2+}2x-rac{2}{12-k}=0 and applying the discriminant formula $D=b^{2-}4ac=0$ yields 4-4(1)(-rac{2}{12-k})=0, which simplifies to 1+rac{2}{12-k}=0 and provides $k=14$. Substituting this value into the coordinates of the point results in $(14, 7)$, and using the formula for the perpendicular distance from a point $(x_1, y_1)$ to the line $3x+4y+5=0$, which is $\frac{|Ax_1+By_1+C|}{\sqrt{A^{2+}B^2}}$, we calculate $\frac{|3(14)+4(7)+5|}{\sqrt{3^{2+}4^2}}$, which equals 15.

Power sums of roots $\alpha, \beta$ of a polynomial $ax^{2+}bx+c=0$ satisfy the relation $aP_{n+2}+bP_{n+1}+cP_n=0$, allowing higher-order terms to be reduced. For the roots $\gamma, \delta$ of $x^{2+}3x-1=0$, this implies $\gamma^{2-}1=-3\gamma$, transforming the expression $\frac{Q_{25}-Q_{23}}{Q_{24}}$ into $\frac{\gamma^{23}(\gamma^{2-}1)+\delta^{23}(\delta^{2-}1)}{Q_{24}} = \frac{-3(\gamma^{24}+\delta^{24})}{Q_{24}} = -3$. Similarly, for the roots $\alpha, \beta$ of $x^{2+}\sqrt{3}x-16=0$, the relation $\alpha^{2+}\sqrt{3}\alpha=16$ leads to $\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}} = \frac{\alpha^{23}(\alpha^{2+}\sqrt{3}\alpha)+\beta^{23}(\beta^{2+}\sqrt{3}\beta)}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$. Combining these results yields $8-3=5$.

Using Vieta's formulas for the equation 2x^2 + (cos θ)x - 1 = 0, we identify the sum of the roots as α+β = -cos θ / 2 and their product as αβ = -1/2. The expression α^{4+}β^4 can be rewritten as (α^{2+}β^2)^2 - 2α^2β^2, which simplifies to (cos^2 θ / 4 + 1)^2 - 1/2. Since cos^2 θ ranges from 0 to 1, the maximum value M occurs when cos^2 θ = 1, giving M = 17/16, and the minimum value m occurs when cos^2 θ = 0, giving m = 1/2. Therefore, 16(M+m) = 16(17/16 + 8/16) = 17 + 8 = 25.

Recurrence relations for power sums $P_n$ of roots $\alpha$ and $\beta$ are defined by the characteristic quadratic equation $x^{2-}(\alpha+\beta)x+\alpha\beta=0$. Given $P_{10}=P_9+P_8$, the roots satisfy $x^{2-}x-1=0$, implying $\alpha+\beta=1$ and $\alpha\beta=-1$. To find the equation for roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$, identify the new sum $\frac{\alpha+\beta}{\alpha\beta} = -1$ and the new product $\frac{1}{\alpha\beta} = -1$, resulting in the equation $x^{2-}(-1)x+(-1)=0$, or $x^{2+}x-1=0$.

Let $t = x^2 - 9x + 11$, noting that $(x-4)(x-5) = x^2 - 9x + 20$ can be written as $t+9$. The equation simplifies to $t^2 - (t+9) = 3$, or $t^2 - t - 12 = 0$, which factors as $(t-4)(t+3) = 0$. Solving $x^2 - 9x + 11 = 4$ yields irrational roots, while solving $x^2 - 9x + 11 = -3$ yields $x^2 - 9x + 14 = 0$, which factors to $(x-7)(x-2)=0$. The product of these two rational roots $7$ and $2$ is $14$.

A polynomial of degree 2 satisfying the functional equation $f(x)f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$ takes the form $f(x) = 1 - x^2$ upon equating coefficients. Setting $f(K) = -2K$ leads to $1 - K^2 = -2K$, which rearranges to the quadratic $K^2 - 2K - 1 = 0$. Letting the roots be $\alpha$ and $\beta$, we apply the relation $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$, yielding $(2)^2 - 2(-1) = 6$.

The equation $x^{2+}3x+2=\min\{|x-3|,|x+2|\}$ represents the intersection of the parabola $y = (x+\frac{3}{2})^2 - \frac{1}{4}$ and the graph of $y = \min\{|x-3|,|x+2|\}$. Since the vertex of the parabola is at $(\frac{-3}{2}, \frac{-1}{4})$ and the minimum of the two absolute value functions is always non-negative, the parabola crosses the branches of the absolute value functions at two distinct points. A graphical analysis confirms these intersection points correspond to two real solutions.

The absolute value expression $|2x-3|$ changes behavior at the critical point $x = \frac{3}{2}$. For $x \ge \frac{3}{2}$, the equation $x^2 + (2x-3) - 4 = 0$ simplifies to $x^2 + 2x - 7 = 0$, giving the valid root $x = 2\sqrt{2}-1$. For $x < \frac{3}{2}$, the equation $x^2 - (2x-3) - 4 = 0$ simplifies to $x^2 - 2x - 1 = 0$, giving the valid root $x = 1-\sqrt{2}$. Summing the squares of these two roots, $(1-\sqrt{2})^2 + (2\sqrt{2}-1)^2$, results in $6(2-\sqrt{2})$.

The quadratic equation $x^{2+}4x-n=0$ has roots x = -2 eq ar{4+n}. For these to be integers, $4+n$ must be a perfect square $k^2$ where $k>0$. Given $n eq [20, 100]$, the value $4+n$ must be \in the range $[24, 104]$. The perfect squares \in this range are $25, 36, 49, 64, 81, 100$, leading to $n$ values of $21, 32, 45, 60, 77, 96$, which amounts to $6$ distinct values.

We determine the number of distinct real roots by solving the equation |x+1||x+3| - 4|x+2| + 5 = 0 across intervals defined by the critical points x=-3, -2, -1. For x ≤ -3, the equation simplifies to (x+4)^2 = 0, yielding x = -4. For x ≥ -1, the equation simplifies to x^2 = 0, yielding x = 0. Testing the intermediate intervals -3 < x ≤ -2 and -2 < x < -1 reveals no valid roots, confirming there are exactly 2 distinct real solutions.

Given the quadratic equation x^2 + 2√2x - 1 = 0, Vieta's formulas give α+β = -2√2 and αβ = -1. We calculate α^{4+}β^4 = ((α+β)^2 - 2αβ)^2 - 2(αβ)^2 = 10^2 - 2 = 98, and α^{6+}β^6 = (α^{3+}β^3)^2 - 2(αβ)^3 = (-22√2)^2 + 2 = 970. The second root is 970/10 = 97, so the quadratic equation with roots 98 and 97 is x^2 - (98+97)x + (98 × 97) = 0, which is x^2 - 195x + 9506 = 0.

By applying Newton's sum relation $a_{n+2} - (t^2 - 5t + 6)a_{n+1} + a_n = 0$ to the given quadratic equation, we can express $a_{2025} + a_{2023}$ as $(t^2 - 5t + 6)a_{2024}$. This simplifies the ratio $\frac{a_{2025} + a_{2023}}{a_{2024}}$ to the quadratic expression $t^2 - 5t + 6$. Completing the square for this function results in $(t - 2.5)^2 - 0.25$, which has a minimum value of $-\frac{1}{4}$ at $t = 2.5$.

Relate the sum and product of roots to the coefficients using Vieta's formulas, where for $px^{2+}qx-r=0$, $\alpha+\beta=-q/p$ and $\alpha\beta=-r/p$. Since $p, q, r$ form a geometric progression, we can substitute $q=pr/p$ to find that the equation simplifies to $x^{2+}Rx-R^2=0$ by letting $R$ be the common ratio. Given $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{1}{R}=\frac{3}{4}$, we determine $R=4/3$. Substituting this into (\alpha-\beta)^2=(\alpha+\eta)^{2-}4\alpha\beta=R^{2+}4R^2=5R^2 gives $5(16/9)=80/9$.

Determine the roots by analyzing the equation within distinct intervals defined by the critical points $x=0$, $x=-1$, and $x=-2$. For $x \ge 0$, the equation simplifies to $x(x+2)-5(x+1)-1=0$, which becomes $x^{2-}3x-6=0$, yielding one positive solution. In the interval $-1 \le x < 0$, the equation becomes $-x(x+2)-5(x+1)-1=0$, leading to $x^{2+}7x+6=0$ and providing $x=-1$ as a solution. For $-2 \le x < -1$, the expression yields $-x(x+2)+5(x+1)-1=0$, simplifying to $x^{2-}3x-4=0$, which contributes no valid roots within that range. Finally, for $x < -2$, the equation becomes $x(x+2)+5(x+1)-1=0$, or $x^{2+}7x+4=0$, which yields one additional valid root, resulting in a total of 3 distinct real roots.

Given the polynomial $P(x) = 4(x - x_1)(x - x_2)(x - x_3)(x - x_4)$, the expression $(4 + x_1^2)(4 + x_2^2)(4 + x_3^2)(4 + x_4^2)$ can be expressed as the product of pairs $\prod (x_k - 2i)(x_k + 2i)$. This product is equivalent to $\frac{P(2i)}{4} \cdot \frac{P(-2i)}{4}$. Evaluating $P(2i)$ by substituting $2i$ into the polynomial results in $141 - 88i$, and $P(-2i)$ is its conjugate $141 + 88i$. The product of these values divided by 16 is $\frac{141^2 + 88^2}{16} = \frac{27625}{16}$. Equating this to the given expression $\frac{125}{16}m$ allows us to solve for $m$ as $27625 / 125$, which is 221.

The inequality holds for all real x if the denominator is always positive and the numerator is always negative. Since the discriminant of the denominator x^2 - 8x + 32 is negative and its leading coefficient is positive, it is always positive. For the numerator to be negative for all x, its leading coefficient a must be negative and its discriminant D = [2(a+1)]^2 - 4a(9a+4) must be less than zero, which simplifies to a < -1/2, yielding no positive integral values for a.

The roots of the equation x^2 - √2x - √3 = 0 satisfy the recurrence relation P_{n+2} = √2P_{n+1} + √3P_n, where P_n = α^n - β^n. By substituting the relation P_{12} = √2P_{11} + √3P_{10} into the given expression (11√3 - 10√2)P_{10} + (11√2 + 10)P_{11} - 11P_{12}, the expression simplifies to 10P_{11} - 10√2P_{10}. Factoring out 10 and applying the recurrence relation P_{11} - √2P_{10} = √3P_9, the final result is 10√3P_9.