Quadratic Equation And Inequalities – JEE Main MathematicsPractice Questions & PYQs

The discriminant $D = b^2 - 4c$ of a quadratic equation is fundamentally linked to the difference between its roots, which is given by $|\alpha - \beta| = \frac{\sqrt{D}}{|A|}$. Since the roots are consecutive integers, their difference is 1, so $|\alpha - (\alpha+1)| = 1$. Squaring both sides and using the coefficient $A=1$, the equation becomes $D = 1^2$, confirming that the value of $b^2 - 4c$ is 1.

For a quadratic equation $x^2 + Bx + C = 0$, the sum of the squares of the roots $\alpha^2 + \beta^2$ is represented by the identity $(\alpha + \beta)^2 - 2\alpha\beta$. Substituting the sum $a-2$ and product $-(a+1)$ from the given equation yields $(a-2)^2 - 2(-(a+1))$, which simplifies to the expression $a^2 - 2a + 6$. Completing the square results in $(a-1)^2 + 5$, which clearly reaches its minimum value when $a-1 = 0$, identifying $a=1$ as the required value.

A quadratic equation is uniquely determined by the sum and product of its roots, taking the form x^2 - (\text{\sum of roots})x + (\text{product of roots}) = 0. Given the arithmetic mean is $9$, the sum of the two numbers is $2 \times 9 = 18$. Since the geometric mean is $4$, the product of the numbers is $4^2 = 16$. Placing these into the standard quadratic form yields $x^2 - 18x + 16 = 0$.

Substituting the given root $4$ into the first equation $x^2 + px + 12 = 0$ gives $16 + 4p + 12 = 0$, which solves to $p = -7$. For a quadratic equation to have equal roots, its discriminant must be zero, expressed as $D = p^2 - 4(1)(q) = 0$. Using the value of $p$ previously found, the equation becomes $(-7)^2 - 4q = 0$, resulting in $49 - 4q = 0$, and finally $q = \frac{49}{4}$.

If $(1-p)$ is a root, substituting it into the equation $x^2 + px + (1-p) = 0$ yields $(1-p)^2 + p(1-p) + (1-p) = 0$. Factoring out $(1-p)$ results in $(1-p)(1 - p + p + 1) = 0$, which simplifies to $2(1-p) = 0$, meaning $p = 1$. With $p = 1$, the original quadratic equation becomes $x^2 + x = 0$, which factors as $x(x + 1) = 0$, providing the roots $0$ and $-1$.

Vieta's formulas establish that the sum of the roots $\alpha + \beta = -b/a$ and the product $\alpha\beta = c/a$. Substituting these into the condition $\alpha + \beta = 1/\alpha^2 + 1/\beta^2$ yields $-b/a = ((\alpha + \beta)^2 - 2\alpha\beta) / (\alpha\beta)^2$. Simplifying this expression leads to $2a^2c = ab^2 + bc^2$, which can be rewritten as $2(a/b) = c/a + b/c$ by dividing throughout by $abc$. This identifies that $c/a$, $a/b$, and $b/c$ form an arithmetic progression, confirming that their reciprocals $a/c$, $b/a$, and $c/b$ are in harmonic progression.

By Vieta's formulas, for roots $\alpha$ and $2\alpha$, the sum $3\alpha = -(3a-1)/(a^{2-}5a+3)$ and the product $2\alpha^2 = 2/(a^{2-}5a+3)$. Equating the expressions for $\alpha^2$ derived from both equations, (1-3a)^2 / \[9(a^{2-}5a+3)^2]$ = 1/(a^{2-}5a+3)$, simplifies to$(1-3a)^2/9 = a^{2-}5a+3$. Expanding and rearranging terms results in$39a = 26$, which provides the value$a = 2/3$.

Recognizing that $x^2$ is equivalent to $|x|^2$, the equation can be rewritten as $|x|^2 - 3|x| + 2 = 0$, which is a quadratic in terms of the absolute value of $x$. Factoring this expression gives $(|x| - 2)(|x| - 1) = 0$, leading to the two cases $|x| = 2$ and $|x| = 1$. Solving these absolute value equations yields $x = \pm 2$ and $x = \pm 1$, resulting in four distinct real solutions.

According to Vieta's formulas, the product of the roots for a quadratic equation $ax^2 + bx + c = 0$ is defined as $c/a$. In the equation $t^2 x^2 + |x| + 9 = 0$, the product of the roots corresponds to $9/t^2$. Since the square of any real number is inherently positive, this ratio remains positive regardless of the value of $t$.

The squared difference between roots of $x^2 + px + q = 0$ is given by $p^2 - 4q$. Setting these differences equal for both equations leads to the equality $a^2 - 4b = b^2 - 4a$. Rearranging terms yields $a^2 - b^2 + 4a - 4b = 0$, which factors into $(a - b)(a + b + 4) = 0$. Because the condition specifies $a \neq b$, the term $a + b + 4$ must be zero.

The values $\alpha$ and $\beta$ satisfy $x^{2-}5x+3=0$, implying the sum of roots $\alpha+\beta=5$ and product $\alpha\beta=3$ from Vieta's formulas. Forming a new equation with roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$, the new sum is $\frac{\alpha^{2+}\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^{2-}2\alpha\beta}{\alpha\beta} = \frac{25-6}{3} = \frac{19}{3}$ and the new product is $\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1$. This results in $x^2 - \frac{19}{3}x + 1 = 0$, which simplifies to $3x^{2-}19x+3=0$.

Vieta's formulas dictate that for the equation $x^{2+}px+q=0$, the sum of the roots is $p+q=-p$ and the product is $pq=q$. Rearranging the product equation gives $q(p-1)=0$, which implies either $q=0$ or $p=1$. Testing $p=1$ in the sum equation yields $1+q=-1$, resulting in $q=-2$.

The algebraic identity $(a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2 + b^2 + c^2 - (ab + bc + ca))$ is essential here because the square of any non-zero real number is positive. Given that $a, b,$ and $c$ are distinct, the sum of these squared differences is greater than zero, which implies $a^2 + b^2 + c^2 > ab + bc + ca$. Since the problem defines $a^2 + b^2 + c^2 = 1$, it follows that $ab + bc + ca$ must be strictly less than 1.

For any quadratic equation $x^{2-}Sx+P=0$, the sum of the roots is $S = \alpha+\beta+\frac{1}{\alpha}+\frac{1}{\beta}$ and the product is P = (\alpha+\frac{1}{\eta})(\beta+\frac{1}{\alpha}). Given the original equation $x^{2-}5x-3=0$ has roots with a sum of $5$ and a product of $-3$, we compute the new sum as $S = 5 + \frac{5}{-3} = \frac{10}{3}$ and the new product as $P = -3 + 2 - \frac{1}{3} = -\frac{4}{3}$. This constructs the quadratic $x^{2-}\frac{10}{3}x-\frac{4}{3}=0$, which simplifies to $3x^{2-}10x-4=0$.

Given the roots α and β of x^2 - 70x + λ = 0, we know α + β = 70 and αβ = λ. Since neither 2 nor 3 divides λ, α and β cannot be multiples of 2 or 3, leading to the pair α = 5 and β = 65 with λ = 325. Substituting these into (√(α-1) + √(β-1))(λ+35)/|α - β| yields (√4 + √64)(360)/60, which equals 60.

The roots of the equation $x^{2-}\sqrt{2}x+\sqrt{6}=0$ satisfy $\alpha+\beta=\sqrt{2}$ and $\alpha\beta=\sqrt{6}$, which allows for the calculation of the roots of the new equation through the transformation $y=\frac{1}{\alpha^2}+1$. Summing these roots yields $a = -\frac{7-\sqrt{6}}{3}$ and their product yields $b = \frac{9-2\sqrt{6}}{6}$. Substituting these into the secondary equation $x^{2-}(a+b-2)x+(a+b+2)=0$ results in a quadratic with a positive discriminant and a negative sum of roots, confirming that both roots are real and negative.

Given the roots for $x^{2-}4\lambda x+5=0$ and $x^{2-}(3\sqrt{2}+2\sqrt{3})x+7+3\sqrt{3}\lambda=0$ with common root $\alpha$, we have $\alpha+\beta=4\lambda$ and $\alpha+\gamma=3\sqrt{2}+2\sqrt{3}$. Summing these gives $2\alpha+\beta+\gamma = 4\lambda+3\sqrt{2}+2\sqrt{3}$, and substituting $\beta+\gamma=3\sqrt{2}$ yields $2\alpha=4\lambda+2\sqrt{3}$ or $\alpha=2\lambda+\sqrt{3}$. Plugging $\alpha$ into the first equation leads to $4\lambda^{2-}8=0$, so $\lambda=\sqrt{2}$. The target expression is $(\alpha+\beta+\beta+\gamma)^2 = (4\lambda+3\sqrt{2})^2 = (4\sqrt{2}+3\sqrt{2})^2 = 98$.

For the quadratic equation $(k+1)\tan^2 x - \sqrt{2}\lambda\tan x + (k-1) = 0$, Vieta's formulas give the sum of the roots $\tan \alpha + \tan \beta = \frac{\sqrt{2}\lambda}{k+1}$ and the product $\tan \alpha \tan \beta = \frac{k-1}{k+1}$. Applying the tangent addition formula, $\tan(\alpha+\beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$, the expression simplifies to $\frac{\sqrt{2}\lambda/(k+1)}{1 - (k-1)/(k+1)} = \frac{\sqrt{2}\lambda}{2} = \frac{\lambda}{\sqrt{2}}$. Given $\tan^2(\alpha+\beta) = 50$, we square this result to get $\frac{\lambda^2}{2} = 50$, which leads to $\lambda^2 = 100$, confirming $\lambda = 10$.