NIMCET 2027 Quadratic Equations Previous Year Questions (PYQs) – Page 3 of 3
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If $\alpha$ and $\beta$ are the roots of the equation $2x^{2}+ 2px + p^{2} = 0$, where $p$ is a non-zero real number, and $\alpha^{4}$ and $\beta^{4}$ are the roots of $x^{2} - rx + s = 0$, then the roots of $2x^{2} - 4p^{2}x + 4p^{4} - 2r = 0$ are:
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Solution
If x and y are positive real numbers satisfying the system of equations $x^{2}+y\sqrt{xy}=336$ and $y^{2}+x\sqrt{xy}=112$, then x + y is:
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Solution
$x^2+y\sqrt{xy}=336 \quad ...(1)$
$y^2+x\sqrt{xy}=112 \quad ...(2)$
Dividing (1) by $\sqrt{x}$,
$x\sqrt{x}+y\sqrt{y}=\frac{336}{\sqrt{x}}$
Dividing (2) by $\sqrt{y}$,
$y\sqrt{y}+x\sqrt{x}=\frac{112}{\sqrt{y}}$
Since LHS are equal,
$\frac{336}{\sqrt{x}}=\frac{112}{\sqrt{y}}$
$\frac{\sqrt{x}}{\sqrt{y}}=3$
$\sqrt{x}=3\sqrt{y}$
$\therefore x=9y$
Substituting in (2),
$y^2+9y\sqrt{9y^2}=112$
$y^2+27y^2=112$
$28y^2=112$
$y^2=4$
$\therefore y=2$
$x=9(2)=18$
$\therefore \boxed{x+y=20}$
The value of k for which the equation $(k-2)x^{2}+8x+k+4=0$
has both real, distinct and
negative roots is
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Solution
For the quadratic equation $(k-2)x^2 + 8x + (k+4) = 0$ to have real, distinct, and negative roots:
1. Real & distinct roots ($D > 0$):
$D = 8^2 - 4(k-2)(k+4) > 0$
$64 - 4(k^2 + 2k - 8) > 0 \implies k^2 + 2k - 24 < 0$
$(k+6)(k-4) < 0 \implies -6 < k < 4$
2. Both roots negative (Sum < 0 & Product > 0):
Sum$ = -\frac{8}{k-2} < 0 \implies k - 2 > 0 \implies k > 2$
* Product $ = \frac{k+4}{k-2} > 0 \implies k \in (-\infty, -4) \cup (2, \infty)$
Taking the intersection of all conditions:
$k \in (2, 4) \quad \text{or} \quad 2 < k < 4$
Roots of equation are $ax^2-2bx+c=0$ are n and m ,
then the value of $\frac{b}{an^2+c}+\frac{b}{am^2+c}$ is
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Solution
Given quadratic equation is $ax^2 - 2bx + c = 0$.
Since $n$ and $m$ are the roots of this equation:
$an^2 - 2bn + c = 0 \implies an^2 + c = 2bn$
$am^2 - 2bm + c = 0 \implies am^2 + c = 2bm$
Now, the given expression is:
$\frac{b}{an^2+c} + \frac{b}{am^2+c}$
Substitute $an^2 + c = 2bn$ and $am^2 + c = 2bm$:
$= \frac{b}{2bn} + \frac{b}{2bm}$
Canceling $b$ from numerator and denominator:
$= \frac{1}{2n} + \frac{1}{2m}$
$= \frac{1}{2} \left( \frac{n+m}{nm} \right)$
From Vieta's formulas:
Sum of roots, $n + m = \frac{-(-2b)}{a} = \frac{2b}{a}$
Product of roots, $nm = \frac{c}{a}$
Substituting these values:
$= \frac{1}{2} \left( \frac{\frac{2b}{a}}{\frac{c}{a}} \right)$
$= \frac{1}{2} \left( \frac{2b}{c} \right)$
$= \frac{b}{c}$
Therefore, the value is $\frac{b}{c}$.
The value of a, for which the sum of the square of the roots of the equation $x^2 - (a -
2)x - (a +1) = 0$, assumes the least value is
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Solution
The given equation is:
$x^2 - (a - 2)x - (a + 1) = 0$
Let the roots be $\alpha$ and $\beta$.
$\alpha + \beta = a - 2$
$\alpha\beta = -(a + 1)$
Let $f(a)$ represent the sum of the squares of the roots:
$f(a) = \alpha^2 + \beta^2$
$f(a) = (\alpha + \beta)^2 - 2\alpha\beta$
Substitute the expressions:
$f(a) = (a - 2)^2 - 2\big(-(a + 1)\big)$
$f(a) = (a^2 - 4a + 4) + (2a + 2)$
$f(a) = a^2 - 2a + 6$
for mininum
$f'(a)=0$
$2a-2=0$
$a=1$
If a + b + c = 0, then the value of $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}$
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Solution
Just Let a=1, b=1 and c=-2 and put in given question
$\frac{1}{-2}+\frac{1}{-2}+\frac{4}{1}$
$\frac{1}{-2}+\frac{1}{-2}+\frac{4}{1}$
$=-1+4=3$
$a, b, c$ are positive integers such that $a^{2}+2b^{2}-2bc=100$ and $2ab-c^{2}=100$. Then the value of $\frac{a+b}{c}$ is
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Solution
Given the two equations:
(1) $a^2 + 2b^2 - 2bc = 100$
(2) $2ab - c^2 = 100$
Equating equation (1) and equation (2):
$a^2 + 2b^2 - 2bc = 2ab - c^2$
Rearranging all terms to one side:
$a^2 - 2ab + 2b^2 - 2bc + c^2 = 0$
Split $2b^2$ into $b^2 + b^2$:
$(a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) = 0$
$(a - b)^2 + (b - c)^2 = 0$
Since $a, b, c$ are real numbers (positive integers), the sum of two non-negative squares can only be zero if each term is zero:
$a - b = 0 \implies a = b$
$b - c = 0 \implies b = c$
Thus:
$a = b = c$
Substitute $a = b = c$ into equation (2):
$2(c)(c) - c^2 = 100$
$2c^2 - c^2 = 100$
$c^2 = 100$
Since $c$ is a positive integer:
$c = 10 \implies a = b = c = 10$
Now, compute the required value:
$\frac{a + b}{c} = \frac{10 + 10}{10} = \frac{20}{10} = 2$
If $x^{2} + 2ax + 10 - 3a > 0$ for all x ∈ R, then
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Solution
Given quadratic inequality:
$x^2 + 2ax + (10 - 3a) > 0 \quad \text{for all } x \in \mathbb{R}$
Condition for a Quadratic to be Strictly Positive:
For any quadratic expression
$Ax^2 + Bx + C > 0$ for all $x \in \mathbb{R}$:
1. $A > 0$ (parabola opens upwards)
2. $D = B^2 - 4AC < 0$ (the parabola does not touch or intersect the x-axis)
Here:
$A = 1 > 0$ (satisfied)
$B = 2a$
$C = 10 - 3a$
Applying the Discriminant Condition ($D < 0$):
$D = (2a)^2 - 4(1)(10 - 3a) < 0$
$4a^2 - 4(10 - 3a) < 0$
Divide both sides by $4$:
$a^2 - (10 - 3a) < 0$
$a^2 + 3a - 10 < 0$
Factor the quadratic equation in $a$:
$(a + 5)(a - 2) < 0$
Using the wavy curve method / sign scheme:
$-5 < a < 2$
or in interval notation:
$a \in (-5, 2)$
Let $f(x) = x^2 - bx + c$, b is an odd positive integer. If f(x)=0 has two prime numbers as roots and b + c = 35, then the global minimum value of f(x) is
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Solution
Given: $f(x) = x^2 - bx + c$
where $b$ is an odd positive integer and $b + c = 35$
Let $\alpha, \beta$ be two prime roots of $f(x) = 0$
By Vieta's formulae:
$\alpha + \beta = b$
$\alpha \beta = c$
Now, $b + c = 35$
$\Rightarrow (\alpha + \beta) + \alpha\beta = 35$
Since $b$ is odd and $\alpha + \beta = b$ is odd
$\Rightarrow$ one of $\alpha, \beta$ must be even prime
$\therefore \alpha = 2$ (only even prime)
$\Rightarrow 2 + \beta + 2\beta = 35$
$\Rightarrow 3\beta = 33$
$\Rightarrow \beta = 11$
$\therefore b = \alpha + \beta = 2 + 11 = 13$
$\therefore c = \alpha\beta = 2 \times 11 = 22$
Verification: $b + c = 13 + 22 = 35$ ✓
So $f(x) = x^2 - 13x + 22$
Global minimum occurs at $x = \dfrac{b}{2} = \dfrac{13}{2}$
Minimum value $= f\left(\dfrac{13}{2}\right) = \left(\dfrac{13}{2}\right)^2 - 13\left(\dfrac{13}{2}\right) + 22$
$= \dfrac{169}{4} - \dfrac{169}{2} + 22$
$= \dfrac{169 - 338 + 88}{4}$
$= \dfrac{-81}{4}$
$\therefore \boxed{\text{Global Minimum} = -\dfrac{81}{4}}$
where $b$ is an odd positive integer and $b + c = 35$
Let $\alpha, \beta$ be two prime roots of $f(x) = 0$
By Vieta's formulae:
$\alpha + \beta = b$
$\alpha \beta = c$
Now, $b + c = 35$
$\Rightarrow (\alpha + \beta) + \alpha\beta = 35$
Since $b$ is odd and $\alpha + \beta = b$ is odd
$\Rightarrow$ one of $\alpha, \beta$ must be even prime
$\therefore \alpha = 2$ (only even prime)
$\Rightarrow 2 + \beta + 2\beta = 35$
$\Rightarrow 3\beta = 33$
$\Rightarrow \beta = 11$
$\therefore b = \alpha + \beta = 2 + 11 = 13$
$\therefore c = \alpha\beta = 2 \times 11 = 22$
Verification: $b + c = 13 + 22 = 35$ ✓
So $f(x) = x^2 - 13x + 22$
Global minimum occurs at $x = \dfrac{b}{2} = \dfrac{13}{2}$
Minimum value $= f\left(\dfrac{13}{2}\right) = \left(\dfrac{13}{2}\right)^2 - 13\left(\dfrac{13}{2}\right) + 22$
$= \dfrac{169}{4} - \dfrac{169}{2} + 22$
$= \dfrac{169 - 338 + 88}{4}$
$= \dfrac{-81}{4}$
$\therefore \boxed{\text{Global Minimum} = -\dfrac{81}{4}}$
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