Quadratic Equation And Inequalities – JEE Main MathematicsPractice Questions & PYQs

The condition for two quadratic equations to share a common root is that the root must satisfy both, which allows us to eliminate the quadratic term. Subtracting the two given equations, $x^{2+}bx-1=0$ and $x^{2+}x+b=0$, yields $x(b-1)-(1+b)=0$, leading to $x=\frac{b+1}{b-1}$. Substituting this expression for $x$ back into $x^{2+}x+b=0$ results in $(\frac{b+1}{b-1})^2 + \frac{b+1}{b-1} + b = 0$, which simplifies to $(b+1)^2 + (b+1)(b-1) + b(b-1)^2 = 0$. Expanding and simplifying, we obtain $b^{3+}3b=0$, which implies $b(b^{2+}3)=0$. Since $b=0$ results in a common root of $-1$, which is excluded, we find $b^2=-3$, yielding $|b|=\sqrt{|-3|}=\sqrt{3}$.

For an equation of the form $A^B=1$, solutions occur when the exponent $B$ is zero (provided the base $A \neq 0$), the base $A$ is $1$, or the base $A$ is $-1$ and the exponent $B$ is an even integer. Setting the exponent $x^{2+}4x-60=0$ yields $x=6$ and $x=-10$. Setting the base $x^{2-}5x+5=1$ gives $x^{2-}5x+4=0$, resulting in $x=1$ and $x=4$. Setting the base $x^{2-}5x+5=-1$ results in $x^{2-}5x+6=0$, giving $x=2$ and $x=3$; testing these, $x=2$ produces an even exponent while $x=3$ produces an odd exponent, making $x=2$ the only valid solution from this case. Summing the valid solutions $1, 4, -10, 6,$ and $2$ gives $3$.

Isolating one radical yields $\sqrt{2x+1} = 1 + \sqrt{2x-1}$. Squaring both sides results in $2x+1 = 1 + (2x-1) + 2\sqrt{2x-1}$, which simplifies to $1 = 2\sqrt{2x-1}$. Squaring again gives $1 = 4(2x-1)$, leading to $x = 5/8$. Substituting this value into $\sqrt{4x^{2-}1}$ yields $\sqrt{4(25/64) - 1} = \sqrt{25/16 - 16/16} = \sqrt{9/16} = 3/4$

For the quadratic equation $x^{2-}6x-2=0$, the roots $\alpha$ and $\beta$ satisfy the recurrence relation $x^n-6x^{n-1}-2x^{n-2}=0$. Multiplying by $\alpha^{n-2}$ and $\beta^{n-2}$ and subtracting the resulting equations yields $a_n-6a_{n-1}-2a_{n-2}=0$, where $a_n=\alpha^n-\beta^n$. This can be rearranged as $a_n-2a_{n-2}=6a_{n-1}$. Dividing both sides by $2a_{n-1}$ results in $\frac{a_n-2a_{n-2}}{2a_{n-1}}=3$. Setting $n=10$ directly provides the value $\frac{a_{10}-2a_8}{2a_9}=3$.

Let $t = \{x\} = x-\lfloor x \rfloor$, where $t \in [0, 1)$. The given equation transforms into $3t^{2-}2t+a^2=0$, or $a^2=3t^{2-}2t$. The function $f(t)=3t^{2-}2t$ has a range of $[-1/3, 1)$ for $t \in [0, 1)$. Since $a$ is a real number, $a^2 \ge 0$, limiting the valid range of $a^2$ to $[0, 1)$. At $a^2=0$, the equation yields $t=0$, which corresponds to an integral solution $x$, which we must exclude. Therefore, for no integral solution to exist, $a^2$ must lie in the interval $(0, 1)$, meaning $a \in (-1, 0) \cup (0, 1)$.

Given $\frac{1}{\alpha}+\frac{1}{\beta}=4$, we use Vieta's formulas to write $\frac{\alpha+\beta}{\alpha\beta} = \frac{-q/p}{r/p} = -q/r = 4$, which implies $q=-4r$. Using the condition that $p, q, r$ are in arithmetic progression, $2q=p+r$, substituting $q=-4r$ leads to $-8r=p+r$, or $p=-9r$. The quadratic equation becomes $-9rx^{2-}4rx+r=0$, which simplifies to $9x^{2+}4x-1=0$. The roots are $\frac{-4 \pm \sqrt{16-4(9)(-1)}}{2(9)} = \frac{-4 \pm \sqrt{52}}{18} = \frac{-2 \pm \sqrt{13}}{9}$, and their absolute difference is $|\frac{-2+\sqrt{13}}{9} - \frac{-2-\sqrt{13}}{9}| = \frac{2\sqrt{13}}{9}$.

If two quadratic equations with real coefficients share a complex root, they must share the conjugate pair, meaning the equations are proportional. For the equations $x^{2+}2x+3=0$ and $ax^{2+}bx+c=0$, this implies the coefficients must maintain the ratio rac{a}{1} = rac{b}{2} = rac{c}{3}. Thus, the ratio $a:b:c$ simplifies to $1:2:3$.

Substituting $u = e^{\sin x}$ into the equation yields $u - \frac{1}{u} - 4 = 0$, which rearranges to the quadratic $u^2 - 4u - 1 = 0$. Solving for $u$ using the quadratic formula gives $u = 2 \pm \sqrt{5}$. Since $e^{\sin x}$ must be positive and lie within the interval $[e^{-1}, e]$, and $2-\sqrt{5} < 0$ while $2+\sqrt{5} \approx 4.23 > e$, no real value of $x$ satisfies either case. Therefore, the equation has no real roots.

The roots of $x^{2-}x+1=0$ are $e^{i\pi/3}$ and $e^{-i\pi/3}$, which satisfy the relation $x^3 = -1$. Because $\alpha^3 = -1$ and $\beta^3 = -1$, the expression $\alpha^{2009} + \beta^{2009}$ becomes $(\alpha^3)^{669}\alpha^2 + (\beta^3)^{669}\beta^2$, which simplifies to $-(\alpha^2 + \beta^2)$. Utilizing the identity $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (1)^2 - 2(1) = -1$, the final result is $-(-1) = 1$.

If the quadratic $bx^{2+}cx+a=0$ has imaginary roots, then $c^{2-}4ab < 0$, which implies $c^2 < 4ab$. The expression $y = 3b^2x^2 + 6bcx + 2c^2$ is a quadratic in $x$ with a minimum value of $-\frac{D}{4A} = -\frac{(6bc)^2 - 4(3b^2)(2c^2)}{4(3b^2)} = -c^2$. Since $c^2 < 4ab$ translates to $-c^2 > -4ab$, it follows that $y \ge -c^2 > -4ab$, meaning the expression is always greater than $-4ab$.

Assigning $\alpha$ as the common root, the roots of $x^{2-}6x+a=0$ are $\alpha$ and $4\beta$, while the roots of $x^{2-}cx+6=0$ are $\alpha$ and $3\beta$. From Vieta's formulas, $\alpha+4\beta=6$ and $\alpha(3\beta)=6$, leading to $\alpha\beta=2$, or $\beta=2/\alpha$. Substituting this into $\alpha+4\beta=6$ gives $\alpha+8/\alpha=6$, which rearranges to $\alpha^{2-}6\alpha+8=0$ and factors to $(\alpha-2)(\alpha-4)=0$. Checking $\alpha=4$ results in non-integer values for the other roots, whereas $\alpha=2$ yields $\beta=1$, making the roots integers, so the common root is $2$.

To evaluate the inequality in the first statement, observe that for any natural number $k$ where $1 \leq k < n$, the term $\frac{1}{\sqrt{k}} > \frac{1}{\sqrt{n}}$. Summing these $n$ individual terms yields $\sum_{k=1}^n \frac{1}{\sqrt{k}} > \sum_{k=1}^n \frac{1}{\sqrt{n}} = n \cdot \frac{1}{\sqrt{n}} = \sqrt{n}$, which confirms Statement 1 is true. The second statement is also true because the inequality $\sqrt{n} < \sqrt{n+1}$ can be rearranged as $\sqrt{n(n+1)} < n+1$, which provides the underlying logic required to prove the bound in Statement 1.

The absolute difference between the roots of a quadratic equation $Ax^2 + Bx + C = 0$ is expressed by the formula $\frac{\sqrt{B^2 - 4AC}}{|A|}$. Substituting the coefficients $A=1$, $B=a$, and $C=1$ from the given equation, the condition for the difference being less than $\sqrt{5}$ becomes $\sqrt{a^2 - 4} < \sqrt{5}$. Squaring both sides yields $a^2 - 4 < 5$, which simplifies to $a^2 < 9$, leading to the solution $a \in (-3, 3)$.

Setting the expression equal to $y$ results in the equation $y = \frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$, which can be rearranged into a quadratic equation $3x^2(y-1) + 9x(y-1) + 7y - 17 = 0$. Since $x$ is a real number, the discriminant must satisfy $D = (9(y-1))^2 - 4(3(y-1))(7y-17) \geq 0$. This inequality simplifies to $(y-1)(y-41) \leq 0$, indicating that $y$ lies within the interval $[1, 41]$, so the maximum value is 41.

Factoring the given quadratic equation $x^2 - 2mx + m^2 - 1 = 0$ allows us to rewrite it as $(x-m)^2 - 1 = 0$. This implies $(x-m)^2 = 1$, resulting in the roots $x_1 = m-1$ and $x_2 = m+1$. Given that both roots must reside in the interval $(-2, 4)$, we set $m-1 > -2$ and $m+1 < 4$, which simplifies to $m > -1$ and $m < 3$ respectively, defining the valid range as $-1 < m < 3$.

Using Vieta's formulas for the quadratic equation $x^2 + px + q = 0$, the sum of the roots is $\tan 30^\circ + \tan 15^\circ = -p$ and their product is $\tan 30^\circ \tan 15^\circ = q$. Applying the tangent addition formula $\tan(45^\circ) = \frac{\tan 30^\circ + \tan 15^\circ}{1 - \tan 30^\circ \tan 15^\circ}$, we substitute the known values to get $1 = \frac{-p}{1 - q}$. Rearranging this gives $1 - q = -p$, which implies $q - p = 1$, and evaluating the expression $2 + q - p$ results in $2 + 1 = 3$.

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.

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.

In a right-angled triangle where $\angle R = \frac{\pi}{2}$, the sum of the other two angles is $\frac{\pi}{2}$, implying $\frac{P}{2} + \frac{Q}{2} = \frac{\pi}{4}$. Utilizing the tangent addition formula $\tan \left(\frac{P}{2} + \frac{Q}{2}\right) = \frac{\tan \left(\frac{P}{2}\right) + \tan \left(\frac{Q}{2}\right)}{1 - \tan \left(\frac{P}{2}\right)\tan \left(\frac{Q}{2}\right)}$, we set the expression equal to $\tan \left(\frac{\pi}{4}\right) = 1$. Substituting the coefficients $-\frac{b}{a}$ for the sum and $\frac{c}{a}$ for the product of roots into this equation yields $\frac{-b/a}{1 - c/a} = 1$, which simplifies to $-b = a - c$, leading to the relation $c = a + b$.

When both roots of a quadratic function are less than a value $p$, the quadratic must satisfy the conditions of a non-negative discriminant for real roots, a vertex position less than $p$, and a positive function value at $p$. Calculating the discriminant $D \ge 0$ results in $4k^2 - 4(k^{2+}k-5) \ge 0$, simplifying to $k \le 5$. The condition $f(5) > 0$ resolves to $k^2 - 9k + 20 > 0$, which is $(k-4)(k-5) > 0$, while the vertex $-b/(2a) = k < 5$. The intersection of $k \le 5$, $k < 4$ or $k > 5$, and $k < 5$ restricts $k$ to the interval $(-\infty , 4)$.