NIMCET Quadratic Equations Previous Year Questions (PYQs) – Page 2 of 3
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The quadratic equation whose roots are $\sin^218^o$ and $\cos^236^o$ is
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Solution
Roots:
$\alpha = \sin^2 18^\circ = \left(\frac{\sqrt{5} - 1}{4}\right)^2 = \frac{3 - \sqrt{5}}{8}$
$\beta = \cos^2 36^\circ = \left(\frac{\sqrt{5} + 1}{4}\right)^2 = \frac{3 + \sqrt{5}}{8}$
Sum of roots ($S$):
$S = \alpha + \beta = \frac{(3 - \sqrt{5}) + (3 + \sqrt{5})}{8} = \frac{6}{8} = \frac{3}{4}$
Product of roots ($P$):
$P = \alpha\beta = \left(\frac{(\sqrt{5} - 1)(\sqrt{5} + 1)}{16}\right)^2 = \left(\frac{4}{16}\right)^2 = \frac{1}{16}$
Equation:
$x^2 - Sx + P = 0$
$x^2 - \frac{3}{4}x + \frac{1}{16} = 0 \implies 16x^2 - 12x + 1 = 0$
The number of roots of the equation
$|x^2-x-6|=x+2$ is
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Solution
$x^2-x-6=(x-3)(x+2)$
Case 1: $x^2-x-6\ge0$
$|x^2-x-6|=x^2-x-6=x+2$
$x^2-2x-8=0 \Rightarrow x=4,-2$
Case 2:
$x^2-x-6<0$
$-(x^2-x-6)=x+2$
$x^2=4 \Rightarrow x=\pm2$
Valid roots: $-2,2,4$
Total roots $=3$
For what value of p, the polynomial $x^4-3x^3+2px^2-6$ is exactly divisible by $(x-1)$
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Solution
Let the given polynomial be:
$P(x) = x^4 - 3x^3 + 2px^2 - 6$
By the Factor Theorem, if $P(x)$ is exactly divisible by $(x - 1)$, then:
$P(1) = 0$
Substitute $x = 1$ into $P(x)$:
$P(1) = (1)^4 - 3(1)^3 + 2p(1)^2 - 6 = 0$
$1 - 3 + 2p - 6 = 0$
$2p - 8 = 0$
$2p = 8$
$p = 4$
Thus, the value of $p$ is $4$.
α, β are the roots of the an equation $x^2- 2x cosθ + 1 = 0$, then the equation having roots αn and βn is
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Solution
The equation $(x-a)^3+(x-b)^3+(x-c)^3=0$ has
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Solution
Let the given equation be:
$f(x) = (x - a)^3 + (x - b)^3 + (x - c)^3 = 0$
Differentiate $f(x)$ with respect to $x$
$f'(x) = 3(x - a)^2 + 3(x - b)^2 + 3(x - c)^2$
$f'(x) = 3\left[(x - a)^2 + (x - b)^2 + (x - c)^2\right]$
Step 2: Analyze the Nature of $f'(x)$
Since the square of any real number is non-negative:
$(x - a)^2 \ge 0, \quad (x - b)^2 \ge 0, \quad (x - c)^2 \ge 0 \quad \text{for all } x \in \mathbb{R}$
If $a, b, c$ are not all equal, the sum of these three non-negative squares cannot simultaneously be zero for any real $x$:
$f'(x) > 0 \quad \text{for all } x \in \mathbb{R}$
If $a = b = c$, $f'(x) \ge 0$ (vanishing only at the single point $x = a$).
In both cases, $f(x)$ is a **strictly increasing function** on $\mathbb{R}$.
Conclusion:
1. As a cubic polynomial with real coefficients, $f(x) \to -\infty$ as $x \to -\infty$, and $f(x) \to +\infty$ as $x \to +\infty$.
2. Because $f(x)$ is strictly increasing, it can cross the x-axis **at most once**.
Therefore, the equation has **exactly one real root** and **two imaginary (complex conjugate) roots**.
Let $a,b,c$ be nonzero real numbers such that $a+b+c\neq 0$ and $4a-2b+c\neq 0$. If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2+bx+c=0$, then which of the following equations has the roots $\frac{\alpha+2}{\alpha-1}$ and $\frac{\beta+2}{\beta-1}$?
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Solution
Let the new root be
$y=\frac{x+2}{x-1}$
Now,
$y(x-1)=x+2$
$xy-y=x+2$
$x(y-1)=y+2$
$x=\frac{y+2}{y-1}$
Since $x$ is a root of
$ax^2+bx+c=0$
Put $x=\frac{y+2}{y-1}$.
$a\left(\frac{y+2}{y-1}\right)^2+b\left(\frac{y+2}{y-1}\right)+c=0$
Multiplying by $(y-1)^2$,
$a(y+2)^2+b(y+2)(y-1)+c(y-1)^2=0$
Now expand:
$a(y^2+4y+4)+b(y^2+y-2)+c(y^2-2y+1)=0$
So,
$(a+b+c)y^2+(4a+b-2c)y+(4a-2b+c)=0$
Therefore, the required equation is
$(a+b+c)x^2+(4a+b-2c)x+(4a-2b+c)=0$
Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - px + r = 0$ and $\dfrac{\alpha}{2}, 2\beta$ are the roots of the equation $x^2 - qx + r = 0$, then the value of $r$,
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Solution
If $\alpha$ and $\beta$ are the two roots of the quadratic equation $x^2 + ax + b = 0, (ab \ne 0)$ then the quadratic roots whose roots $\frac{1}{\alpha^3+\alpha}$ and $\frac{1}{\beta^3+\beta}$ is
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Solution
Given $x^{2} + ax + b = 0$ with roots $\alpha,\beta$,
we use
$\alpha + \beta = -a$ and $\alpha\beta = b$.
Required new roots are
$\dfrac{1}{\alpha^{3}+\alpha}$ and $\dfrac{1}{\beta^{3}+\beta}$.
Since
$\alpha^{3}+\alpha = \alpha(\alpha^{2}+1)$
and using $\alpha^{2}=-a\alpha-b$ (and same for $\beta$),
after simplification the sum and product of new roots become:
$u+v = \dfrac{a^{3}+a-3ab}{b(b^{2}+1+a^{2}-2b)}$
$uv = \dfrac{1}{b(b^{2}+1+a^{2}-2b)}$
So the required quadratic is:
$b(b^{2}+1+a^{2}-2b)x^{2} - (a^{3}+a-3ab)x + 1 = 0$
If α≠β and $\alpha^2=5\alpha-3,\beta^2=5\beta-3$, then the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is
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Solution
Given the equation $x+y =1$, $x^2+y^2 =2$, $x^5 +y^5 =A$. Let N be the number
of solution pairs (x,y) to this system of equations. Then AN is equal to
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Solution
From \(x+y=1\) and \(x^2+y^2=2\): $(x+y)^2=x^2+y^2+2xy$
\[ \;\Rightarrow\; 1=2+2xy \;\Rightarrow\; xy=-\tfrac12.
\]
Thus \(x,y\) are roots of \(t^2 - t - \tfrac12=0\), giving two ordered pairs \((x,y)\). Hence \(N=2\).
Let \(p_n=x^n+y^n\). For quadratic roots with \(s_1=x+y=1\) and \(s_2=xy=-\tfrac12\),
\[
p_n = s_1 p_{n-1} - s_2 p_{n-2} \quad (n\ge2),
\]
with \(p_0=2,\; p_1=1\).
\[
p_2=1\cdot1-(-\tfrac12)\cdot2=2,\quad\]
$$ p_3=1\cdot2-(-\tfrac12)\cdot1=\tfrac52,$$
\[
p_4=\tfrac72,\quad
p_5=1\cdot\tfrac72-(-\tfrac12)\cdot\tfrac52=\tfrac{19}{4}.
\]
Therefore \(A=p_5=\tfrac{19}{4}\). With \(N=2\),
\[
AN=\frac{19}{4}\cdot 2=\boxed{\tfrac{19}{2}}.
\]
✅ Final Answer: \(AN=\dfrac{19}{2}\)
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