Quadratic Equation And Inequalities – JEE Main MathematicsPractice Questions & PYQs

With roots $\alpha$ and $\beta$ for $x^{2-}3x+r=0$, the relations $\alpha+\beta=3$ and $\alpha\beta=r$ hold. For the second equation $x^{2+}3x+r=0$ having roots $\frac{\alpha}{2}$ and $2\beta$, we have $\frac{\alpha}{2}+2\beta=-3$ and $\alpha\beta=r$. Solving the system $\alpha+\beta=3$ and $\alpha+4\beta=-6$ leads to $\beta=-3$ and $\alpha=6$, which gives $r=-18$. The roots of $x^{2+}6x-m=0$ are $2\alpha+\beta+2r = -27$ and $\alpha-2\beta-\frac{r}{2} = 21$, so the product of these roots is $-m = (-27)(21) = -567$, meaning $m=567$.

The roots of an equation involving absolute value terms can be found by substituting $t = \sqrt{x}$ and rewriting the expression $x-6\sqrt{x}$ as $(t-3)^{2-}9$. This simplifies the original equation to $(-2+\sqrt{3})|t-3| + |t-3|^2 - 2\sqrt{3} = 0$. By solving this quadratic for $y = |t-3|$, we obtain $y=2$, implying $\sqrt{x}-3 = \pm 2$, which gives $\sqrt{x} = 5$ or $1$, resulting in $\alpha=1$ and $\beta=25$. The final evaluation of $\sqrt{\frac{25}{1}} + \sqrt{1 \times 25}$ equals $5+5=10$.

A quadratic equation $Ax^{2+}Bx+C=0$ has no real roots when its discriminant $B^{2-}4AC$ is strictly negative. Applying this to $2x^{2+}(a-5)x+(15-3a)=0$ yields $(a-5)^{2-}8(15-3a) < 0$, which simplifies to $a^{2+}14a-95 < 0$ and factors as $(a+19)(a-5) < 0$, defining the interval $(-19, 5)$. The set $X$ of integers in this range is $\{-18, -17, \dots, 4\}$, and summing their squares, computed as $\sum_{x=1}^{18} x^2 + \sum_{x=1}^{4} x^2$, results in $2109 + 30 = 2139$.

For a limit to exist as a finite value, the numerator must approach zero when the denominator approaches zero. Applying the standard limit $\lim_{x\to 2} \frac{\tan(x-2)}{x-2} = 1$, the expression simplifies to the limit of $\frac{rx^{2}+(p-2)x-2p}{x-2}$ as $x \to 2$, which equals 5. Evaluating the numerator at $x=2$ gives $4r+2p-4-2p = 4r-4$, so $r=1$ is necessary for the limit to be defined; subsequent substitution yields $p=3$. The roots of $x^{2-}3x+q=0$ lie in $(0, 2)$ if the discriminant is non-negative ($D = 9-4q \geq 0 \Rightarrow q \leq 2.25$), the vertex $1.5$ lies in $(0, 2)$, and the function at boundaries is positive ($f(0)=q>0$ and $f(2)=q-2>0$), implying $q \in (2, 2.25]$. With $\alpha=2$ and $\beta=2.25$, $4(\alpha+\beta)$ results in 17.

When a polynomial is divisible by a quadratic expression, the remainder of the division must be identically zero. For the polynomial $x^{3+}ax^{2+}bx+c$, substituting $x^2=-2$ must reduce the polynomial to zero. This yields $(b-2)x+(c-2a)=0$, which implies $b=2$ and $c=2a$. Given $a \in \{1, 2, \dots, 20\}$, the condition $c \le 20$ leads to $2a \le 20$, or $a \le 10$, resulting in 10 possible polynomials.

The difference between the roots of a quadratic equation $ax^{2+}bx+c=0$ is given by $|\beta-\alpha| = \frac{\sqrt{b^{2-}4ac}}{|a|}$. Substituting the given coefficients into this formula yields $|\beta-\alpha| = \frac{\sqrt{400-144\lambda}}{12}$. Applying the inequality $\frac{1}{2} \leq \frac{\sqrt{400-144\lambda}}{12} \leq \frac{3}{2}$ results in the range $36 \leq 400-144\lambda \leq 324$, which is satisfied only by the integers $\lambda=1$ and $\lambda=2$. Summing these values gives a total of $3$.

The number of real solutions to an equation involving absolute value terms is determined by examining each interval defined by the roots of the absolute value expressions. Analyzing the equation in the intervals $x \geq -2$, $-4 \leq x < -2$, and $x < -4$ shows that the first two cases result in quadratic forms with negative discriminants, yielding no real roots. In the range $x < -4$, the equation simplifies to $x^{2+}7x-4=0$, which produces exactly one valid solution, $x = \frac{-7 - \sqrt{65}}{2}$, as the other root falls outside the interval.

For a quadratic function $f(x) = x^{2+}2ax+(3a+10)$ to have roots $\alpha$ and $\beta$ such that $\alpha < 1 < \beta$, the function must satisfy $f(1) < 0$ because the parabola opens upwards. Evaluating $f(1) = 1^2 + 2a(1) + 3a + 10$ leads to the expression $5a+11$, and the inequality $5a+11 < 0$ simplifies to $a < -\frac{11}{5}$. This condition ensures that the parabola intersects the x-axis on both sides of $x=1$, guaranteeing two distinct real roots.

For a biquadratic equation of the form $x^{4-}ax^{2+}9=0$, real and distinct roots require the transformed quadratic $y^{2-}ay+9=0$ to have two distinct positive roots. The condition for distinct real roots in $y$ is $D = a^{2-}36 > 0$, implying $a > 6$. Since the product of roots is $9$ and the sum is $a$, both roots are positive if $a > 0$, which is satisfied by $a > 6$. The smallest positive integer satisfying this is $7$.

The relationship between the roots of a quadratic equation and its coefficients enables expressing the given difference of reciprocals in terms of the discriminant and product of roots. For an equation $\lambda x^2 - (\lambda+3)x + 3 = 0$, the roots $\alpha, \beta$ satisfy $\alpha+\beta = \frac{\lambda+3}{\lambda}$ and $\alpha\beta = \frac{3}{\lambda}$. Substituting these into the condition $\frac{\beta-\alpha}{\alpha\beta} = \frac{1}{3}$ yields $\frac{\sqrt{(\alpha+\beta)^{2-}4\alpha\beta}}{\alpha\beta} = \frac{1}{3}$, which simplifies to $\sqrt{\lambda^{2-}6\lambda+9} = 1$. Solving $|\lambda-3|=1$ gives $\lambda=4$ and $\lambda=2$, resulting in a sum of $6$.

The absolute value function necessitates splitting the equation into different intervals defined by the critical points $x=-3$ and $x=1$. In the interval $x < -3$, the equation reduces to $x^{2+}4x+1=0$, yielding $x = -2-\sqrt{3}$, which is a valid solution. For $-3 \le x < 1$, the equation becomes $x^{2+}2x-1=0$, providing two roots $x = -1\pm\sqrt{2}$, both of which fall within the defined interval. For $x \ge 1$, the equation results in $x^{2+}4x-3=0$, which yields no solutions within the range, totaling three real solutions.

A quadratic equation $ax^{2+}bx+c=0$ has two positive roots if the discriminant is non-negative, the product of roots is positive, and the sum of roots is positive. For the given equation, the product $\frac{4\lambda}{\lambda+2} > 0$ and the sum $\frac{3\lambda}{\lambda+2} > 0$ both constrain $\lambda$ to $(-\infty , -2) \cup (0, \infty )$. The discriminant condition $D = 9\lambda^2 - 16\lambda(\lambda+2) = -7\lambda^2 - 32\lambda \geq 0$ further requires $\lambda \in [-\frac{32}{7}, 0]$. Intersecting these conditions restricts $\lambda$ to $[-\frac{32}{7}, -2)$, within which the possible integral values for $\lambda$ are $-4$ and $-3$, totaling 2 values.

The combined rate of two people working together equals the sum of their individual rates, expressed as $\frac{1}{A} + \frac{1}{B} = \frac{1}{T}$. Letting $x$ represent the time taken by mason B, mason A takes $x-24$ days, which leads to the equation $\frac{1}{x-24} + \frac{1}{x} = \frac{1}{22.5} = \frac{2}{45}$. Simplifying this expression to the quadratic equation $x^2 - 69x + 540 = 0$ and solving for $x$ gives $x=60$, which means mason A requires $60-24=36$ days to complete the work.

The triangle inequality property $|a+b| = |a| + |b|$ holds if and only if $ab \geq 0$. Applying this to the equation $|(x^{2-}9) + x| = |x| + |x^{2-}9|$, we must satisfy $x(x^{2-}9) \geq 0$, which factors into $x(x-3)(x+3) \geq 0$. Evaluating this inequality on the number line gives the intervals $[-3, 0] \cup [3, \infty )$. Comparing this to $[\alpha, \beta] \cup [\gamma, \infty )$, we identify $\alpha=-3$, $\beta=0$, and $\gamma=3$, making the sum of their squares $(-3)^2 + 0^2 + 3^2 = 18$.

Substituting $y=|x-1|$ into the expression $(x-1)^{2-}5|x-1|+6=0$ transforms it into $y^{2-}5y+6=0$, which factors as $(y-2)(y-3)=0$. This implies $|x-1|=2$ or $|x-1|=3$. Solving for $x$ leads to $x=3$ and $x=-1$ from the first case, and $x=4$ and $x=-2$ from the second case, giving a total sum of $3+(-1)+4+(-2)=4$.

Summing the terms of the expression $\sum_{k=0}^{n-1} (x+k)(x+k+2)$ yields $nx^2 + n(n+1)x + \frac{n(n-1)(2n+5)}{6} = 4n$. Dividing by $n$ and rearranging gives the quadratic $x^2 + (n+1)x + \frac{(n-1)(2n+5)}{6} - 4 = 0$. Since the roots $\alpha$ and $\alpha+2$ differ by 2, the discriminant must satisfy $\sqrt{D} = 2$, or $D=4$. Solving the resulting equation for $n$ yields $n=7$, leading to the quadratic $x^{2+}8x+15=0$ with roots $-3$ and $-5$; taking $\alpha=-5$, the value of $n+\alpha$ is $7-5=2$.

The domain of a logarithmic function is determined by the constraints that the argument must be strictly positive and the base must be strictly positive and not equal to one. For $f(x)=\log_{(10x^{2-}17x+7)}(18x^{2-}11x+1)$, the argument condition $18x^{2-}11x+1 > 0$ gives $x \in (-\infty , 1/9) \cup (1/2, \infty )$, while the base condition $10x^{2-}17x+7 > 0$ gives $x \in (-\infty , 0.7) \cup (1, \infty )$, and base $\ne 1$ gives $x \ne 0.5, 1.2$. Intersecting these conditions leads to the domain $(-\infty , 1/9) \cup (1/2, 0.7) \cup (1, 1.2) \cup (1.2, \infty )$, identifying $a=1/9, b=1/2, c=7/10, d=1, e=1.2$, which sums to $316$ when multiplied by 90.

For a quadratic equation $x^{2-}(p+2)x+(2p+9)=0$ to have two negative roots, the discriminant $D$ must be at least $0$, the sum of roots must be negative, and the product of roots must be positive. These conditions translate to p eq (-ar{ar{ar{}}} ,-4] eq [8,ar{ar{ar{}}}), $p < -2$, and $p > -4.5$, respectively. Intersecting these conditions gives $p eq (-4.5, -4]$, where ar{ar{ar{}}} = -4.5 and ar{ar{ar{}}} = -4, leading to ar{ar{ar{}}}-2ar{ar{ar{}}} = -4-2(-4.5)=5.

The first equation $|x-2|^{2+}|x-2|-2=0$ factors as $(|x-2|+2)(|x-2|-1)=0$, resulting in $|x-2|=1$ and roots $x=3, 1$. For the second equation $x^{2-}2|x-3|-5=0$, evaluating $x<3$ gives the quadratic $x^{2+}2x-11=0$ with roots x=-1 eq 2ar{3}, while $x eq 3$ yields no valid solutions. Summing the squares of all roots, 3^{2+}1^{2+}(-1+2ar{3})^{2+}(-1-2ar{3})^2, yields 9+1+13-4ar{3}+13+4ar{3}=36.

Analyzing the equation $x|x-2|+3|x-3|+1=0$ across intervals reveals roots only for $x<2$, where the equation becomes $x^{2+}x-10=0$ and yields the root x=rac{-1-ar{41}}{2}. In the interval $2 eq x < 3$, the equation simplifies to $x^{2-}5x+10=0$ with no real roots, and for $x eq 3$, it simplifies to $x^{2+}x-8=0$ with no roots in that range. Consequently, there is only one real root.