COMPLEX NUMBERS AND QUADRATIC EQUATIONS - 1 (Quadratic Equations)

Quadratic equations

The general form of a quadratic equation over real numbers is $a x^{2}+b x+c=0$ where $a, b, c \in R \& a \neq$ 0 . The solution of the quadrati equation $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0$ is given by

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$. The expression $b^{2}-4 a c$ is called the discriminant of the quadratic equation and is denoted by D.

Nature of roots : For the quadratic equation $a x^{2}+b x+c=0$, where $a, b, c \in R$ and $a \neq 0$, then

For the quadratic equation $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0$ where $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{Q}$ and $\mathrm{a} \neq 0$, then :

Note 1: If $\alpha$ is a root of $f(\mathrm{x})=0$, then the polynomial $f(\mathrm{x})$ is exactly divisible by $\mathrm{x}-\alpha$ or $(\mathrm{x}-\alpha)$ is a factor of $f(\mathrm{x})$ and vice versa.

Note 2: $x^{2}+b x+c=0$ cannot have three different roots. If it has, then the equation becomes an identity in $\mathrm{x}$. ie, $\mathrm{a}=\mathrm{b}=\mathrm{c}=0$.

Relation between roots and coefficients

If $\alpha _{1}, \alpha _{2}, \ldots \ldots \ldots \alpha _{n}$ are roots of the equation $f(x)=a _{n} x^{n}+a _{n-1} x^{n-1}+a _{n-2} x^{n-2}+\ldots \ldots \ldots a _{2} x^{2}+a _{1} x+a _{0}=0$, then $f(\mathrm{x})=\mathrm{a} _{\mathrm{n}}\left(\mathrm{x}-\alpha _{1}\right)\left(\mathrm{x}-\alpha _{2}\right)\left(\mathrm{x}-\alpha _{3}\right)$ $\left(\mathrm{x}-\alpha _{\mathrm{n}}\right)$

$\therefore \mathrm{a} _{\mathrm{n}} \mathrm{x}^{\mathrm{n}}+\mathrm{a} _{\mathrm{n}-1} \mathrm{x}^{\mathrm{n}-1}+\mathrm{a} _{\mathrm{n}-2} \mathrm{x}^{\mathrm{n}-2}+\ldots \ldots \ldots \ldots+\mathrm{a} _{2} \mathrm{x}^{2}+\mathrm{a} _{1} \mathrm{x}+\mathrm{a} _{0}=\mathrm{a} _{\mathrm{n}}\left(\mathrm{x}-\alpha _{1}\right)\left(\mathrm{x}-\alpha _{2}\right)$.

Comparing the coefficients of like powers of $x$ an both sides, we get,

$ \begin{aligned} & S _{1}=\alpha _{1}+\alpha _{2}+\ldots \ldots \ldots \ldots+\alpha _{n}=\sum \alpha _{i}=\frac{-a _{n-1}}{a _{n}}=\frac{- \text { coeft. of } x^{n-1}}{\text { coeft of } x^{n}} \\ & S _{2}=\alpha _{1} \alpha _{2}+\alpha _{1} \alpha _{3}+\ldots \ldots \ldots \ldots \ldots \sum _{i \neq j} \alpha _{i} \alpha _{j}=(-1)^{2} \frac{a _{n-2}}{a _{n}}=(-1)^{2} \frac{\text { coeft. of } x^{n-2}}{\text { coeft of } x^{n}} \\ & S _{3}=\alpha _{1} \alpha _{2} \alpha _{3}+\alpha _{2} \alpha _{3} \alpha _{4}+\ldots \ldots \ldots \ldots \ldots \sum _{i \neq j \neq k} \alpha _{i} \alpha _{j} \alpha _{k}=(-1)^{3} \frac{a _{n-3}}{a _{n}}=(-1)^{3} \frac{\text { coeft. of } x^{n-3}}{\text { coeft of } x^{n}} \end{aligned} $

$ \mathrm{S} _{\mathrm{n}}=\alpha _{1} \alpha _{2} \alpha _{3} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \alpha _{n}=(-1)^{\mathrm{n}} \frac{\mathrm{a} _{0}}{\mathrm{a} _{\mathrm{n}}}=(-1)^{\mathrm{n}} \frac{\text { constant term }}{\operatorname{coeft} \text { of } \mathrm{x}^{\mathrm{n}}} $

Here $\mathrm{S} _{\mathrm{k}}$ denotes the sum of the products of the roots taken ’ $\mathrm{k}$ ’ at a time.

Particular cases :-

Quadratic equation : If $\alpha \& \beta$ are the roots of the quadratic equation $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0$, then $\mathrm{S} _{1}=\alpha+\beta=\frac{-\mathrm{b}}{\mathrm{a}}, \& \mathrm{~S} _{2}=\alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}$

Cubic equation: If $\alpha, \beta, \gamma$ are the roots of the cubic equation $\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}=0$, then

$ \begin{aligned} & \mathrm{S} _{1}=\alpha+\beta+\gamma=\frac{-b}{\mathrm{a}} \\ & \mathrm{S} _{2}=\alpha \beta+\beta \alpha+\gamma \alpha=(-1)^{2} \frac{\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{a}} \\ & \mathrm{S} _{3}=\alpha \beta \gamma=(-1)^{3} \frac{\mathrm{d}}{\mathrm{a}}=\frac{-\mathrm{d}}{\mathrm{a}} \end{aligned} $

Biquadratic equation : If $\alpha, \beta, \gamma, \delta$ are roots of the biquadratic equation $\mathrm{ax}^{4}+\mathrm{bx}^{3}+\mathrm{cx}^{2}+\mathrm{dx}+\mathrm{e}=0$, then

$ \begin{aligned} & \mathrm{S} _{1}=\alpha+\beta+\gamma+\delta=\frac{-b}{\mathrm{a}} \quad \frac{\text { constant term }}{\text { coeft of }^{\mathrm{n}}} \\ & \mathrm{S} _{2}=\alpha \beta+\beta \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta=(-1)^{2} \frac{\mathrm{c}}{\mathrm{a}} \\ & \text { Or } \mathrm{S} _{2}=(\alpha+\beta)(\gamma+\delta)+\alpha \beta+\gamma \delta=\frac{\mathrm{c}}{\mathrm{a}} \\ & \mathrm{S} _{3}=\alpha \beta \gamma+\beta \gamma \delta+\gamma \delta \alpha+\alpha \beta \delta=(-1)^{3} \frac{\mathrm{d}}{\mathrm{a}} \\ & \text { Or } \mathrm{S} _{3}=\alpha \beta(\gamma+\delta)+\gamma \delta(\alpha+\beta)=\frac{-\mathrm{d}}{\mathrm{a}} \\ & \text { and } \mathrm{S} _{4}=\alpha \beta \gamma \delta=(-1)^{4} \frac{\mathrm{e}}{\mathrm{a}}=\frac{\mathrm{e}}{\mathrm{a}} \end{aligned} $

Formation of a polynomial equation from given roots

If $\alpha _{1}, \alpha _{2}, \alpha _{3}, \ldots \ldots \ldots \ldots \ldots, \alpha _{n}$ are the roots of an $\mathrm{n}^{\text {th }}$ degree equation, then the equation is $\mathrm{x}^{\mathrm{n}}-\mathrm{S} _{1} \mathrm{x}^{\mathrm{n}-1}+\mathrm{S} _{2} \mathrm{x}^{\mathrm{n}-2}-\mathrm{S} _{3} \mathrm{x}^{\mathrm{n}-3}+\ldots \ldots \ldots \ldots .+(-1)^{\mathrm{n}} \mathrm{S} _{\mathrm{n}}=0$ where $\mathrm{S} _{\mathrm{k}}$ denotes the sum of the products of roots taken $\mathrm{k}$ at a time.

Particular cases

Quadratic equation : If $\alpha, \beta$ are the roots of a quadratic equation, then the equation is $x^{2}-S _{1} x+S _{2}=0$ ie, $x^{2}-(\alpha+\beta) x+\alpha \beta=0$.

Cubic equation : If $\alpha, \beta, \gamma$ are the roots of a cubic equation. Then the equation is, $\mathrm{x}^{3}-\mathrm{S} _{1} \mathrm{x}^{2}+\mathrm{S} _{2} \mathrm{x}-\mathrm{S} _{3}=0$ ie, $\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma=0$

Biquadratic equation : If $\alpha, \beta, \gamma, \delta$ are the roots of a biquadratic equation, then the equation is $\mathrm{x}^{4}-\mathrm{S} _{1} \mathrm{x}^{3}+\mathrm{S} _{2} \mathrm{x}^{2}-\mathrm{S} _{3} \mathrm{x}+\mathrm{S} _{4}=0$

ie, $x^{4}-(\alpha+\beta+\gamma+\delta) x^{3}+(\alpha \beta+\beta \gamma+\gamma \delta+\alpha \delta+\beta \delta+\alpha \gamma) x^{2}-$ $(\alpha \gamma \beta+\alpha \beta \delta+\beta \gamma \delta+\alpha \gamma \delta) x+\alpha \beta \gamma \delta=0$.

Quadratic Expression : An expression of the form $a x^{2}+b x+c$, where $a, b, c \in R \& a \neq 0$ is called a quadratic expression in $\mathrm{x}$. So in general quadratic expression is represented as: $f(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ or $\mathrm{y}=$ $a x^{2}+b x+c$.

Graph of a quadratic Expression

Let $\mathrm{y}=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ where $\mathrm{a} \neq 0$.

Then $y=a\left(x^{2}+\frac{b}{a} x+\frac{c}{a}\right) \Rightarrow y=a\left(x^{2}+\frac{b x}{a}+\frac{b^{2}}{4 a^{2}}+\frac{c}{a}-\frac{b^{2}}{4 a^{2}}\right)$

$\Rightarrow y+\frac{b^{2}-4 a c}{4 a}=a\left(x+\frac{b}{2 a}\right)^{2} \Rightarrow y+\frac{D}{4 a}=a\left(x+\frac{b}{2 a}\right)^{2}$

Let $y+\frac{D}{4 a}=Y \& x+\frac{b}{2 a}=X$

$\therefore \mathrm{Y}=\mathrm{aX}^{2}$ or $\mathrm{X}^{2}=\frac{\mathrm{Y}}{\mathrm{a}}$

Clearly it is the equation of a parabola having its vertex at $\left(\frac{-b}{2 a}, \frac{-D}{4 a}\right)$.

If $\mathrm{a}>0$, then the parabola open upwards.

If $a<0$, then the parabola open downwards.

Sign of quadratic Expression

(1) The parabola will intersect the $x$-axis in two distinct points if $\mathrm{D}>0$.

(i) $ \mathrm{a}>0$

Let $f(\mathrm{x})=0$ have 2 real roots

$\alpha \& \beta(\alpha<\beta)$. Then $f(x)>0$

$\forall \mathrm{x} \in(-\infty, \alpha) \cup(\beta \infty)$ and $f(\mathrm{x})<0$

$\forall \mathrm{x} \in(\alpha, \beta)$

(ii) $ \mathrm{a}<0$

Let $f(\mathrm{x})=0$ have 2 real roots

$\alpha \& \beta(\alpha<\beta)$ Then $f(\mathrm{x})<0$

$\forall \mathrm{x} \in(-\infty, \alpha) \cup(\beta \infty)$

$\& f(\mathrm{x})>0$ for all $\mathrm{x} \in(\alpha, \beta)$

(2) The parabola will touch the $\mathrm{x}$-axis at one point if $\mathrm{D}=0$

(3) The parabola will not intersect $\mathrm{x}$-axis if $\mathrm{D}<0$.

NOTE : Condition that a quadratic function $f(\mathrm{x}, \mathrm{y})=\mathrm{ax}^{2}+2 \mathrm{hxy}+\mathrm{by}^{2}+2 \mathrm{gx}+2 f \mathrm{y}+\mathrm{c}$ may be resolved into two linear factions is that

$a b c+2 f g h-a f^{2}-b^{2}-c h^{2}=\left|\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right|=0$

NOTE :

(i) For $\mathrm{a}>0, f(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ has least value at $\mathrm{x}=\frac{-\mathrm{b}}{2 \mathrm{a}}$. This least value is given by $\frac{-\mathrm{D}}{4 \mathrm{a}}$

(ii) For $\mathrm{a}<0, f(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ has greatest value at $\mathrm{x}=\frac{-\mathrm{b}}{2 \mathrm{a}}$. This greatest value is given by $\frac{-\mathrm{D}}{4 \mathrm{a}}$

Solved Examples

1. If $\alpha, \beta$ are roots of $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0 ; \alpha+\mathrm{h}$ and $\beta+\mathrm{h}$ are roots of $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=0$ and $\mathrm{D} _{1}, \mathrm{D} _{2}$ are their discriminants, then $\mathrm{D} _{1}: \mathrm{D} _{2}=$

(a). $\frac{\mathrm{a}^{2}}{\mathrm{p}^{2}}$

(b). $\frac{b^{2}}{q^{2}}$

(c). $\frac{\mathrm{c}^{2}}{\mathrm{r}^{2}}$

(d). None of these

Show Answer

Solution : $\alpha-\beta=(\alpha+h)-(\beta+h)$

$ \begin{aligned} & \Rightarrow(\alpha-\beta)^{2}=((\alpha+h)-(\beta+h))^{2} \\ & \Rightarrow(\alpha+\beta)^{2}-4 \alpha \beta=((\alpha+h)+(\beta+h))^{2}-4(\alpha+h)(\beta+h) \\ & \left(\frac{-b}{a}\right)^{2}-4 \frac{c}{a}=\left(\frac{-q}{p}\right)^{2}-\frac{4 r}{p} \end{aligned} $

$ \begin{aligned} & \frac{\mathrm{b}^{2}-4 \mathrm{ac}}{\mathrm{a}^{2}}=\frac{\mathrm{q}^{2}-4 \mathrm{pr}}{\mathrm{p}^{2}} \Rightarrow \frac{\mathrm{D} _{1}}{\mathrm{a}^{2}}=\frac{\mathrm{D} _{2}}{\mathrm{p}^{2}} \\ & \Rightarrow \frac{\mathrm{D} _{1}}{\mathrm{D} _{2}}=\frac{\mathrm{a}^{2}}{\mathrm{p}^{2}} \end{aligned} $

Answer : (a).

2. If $a \in Z$ and the equation $(x-a)(x-10)+1=0$ has integral roots, then the values of $a$ are

(a). 10, 8

(b). 12, 10

(c). 12, 8

(d). None of these

Show Answer

Solution: $(\mathrm{x}-\mathrm{a})(\mathrm{x}-10)=-1$

$\begin{aligned} & \Rightarrow \quad x-a=1 & \quad {\&} \quad \quad x-10=-1 \\ & 9-a=1 & x=10-1 \\ & a=8 & x=9 \end{aligned}$

$\begin{aligned} & \text { OR } \quad \mathrm{x}-\mathrm{a}=-1 \quad {\&} & \quad\mathrm{x}-10=1 \\ & 11-a=-1 & x=11 \\ & \mathrm{a}=12 \\ & \end{aligned}$

Answer : (c).

3. If $\alpha, \beta$ are roots of the equation $(x-a)(x-b)+c=0(c \neq 0)$, then then roots of the equation ( $x-c-$ $\alpha)(\mathrm{x}-\mathrm{c}-\beta)=\mathrm{c}$ are

(a). $a$ and $b+c$

(b). $a+\mathrm{c}$ and $b$

(c). $a+c$ and $b+c$

(d). None of these

Show Answer

Solution: $\mathrm{x}^{2}-(\mathrm{a}+\mathrm{b}) \mathrm{x}+\mathrm{ab}+\mathrm{c}=0$

$\Rightarrow \quad \alpha+\beta=\mathrm{a}+\mathrm{b}$ and $\alpha \beta=\mathrm{ab}+\mathrm{c}$

Now $(\mathrm{x}-\mathrm{c}-\alpha)(\mathrm{x}-\mathrm{c}-\beta)=\mathrm{c}$

$\Rightarrow \quad(\mathrm{x}-\mathrm{c})^{2}-(\alpha+\beta)(\mathrm{x}-\mathrm{c})+\alpha \beta-\mathrm{c}=0$

$(\mathrm{x}-\mathrm{c})^{2}-(\mathrm{a}+\mathrm{b})(\mathrm{x}-\mathrm{c})+\mathrm{ab}=0$

$(\mathrm{x}-\mathrm{c})^{2}-\mathrm{a}(\mathrm{x}-\mathrm{c})-\mathrm{b}(\mathrm{x}-\mathrm{c})+\mathrm{ab}=0$

$\Rightarrow \quad(\mathrm{x}-\mathrm{c}-\mathrm{a})(\mathrm{x}-\mathrm{c}-\mathrm{b})=0$

$\therefore \quad \mathrm{x}=\mathrm{c}+\mathrm{a}$ and $\mathrm{b}+\mathrm{c}$

Answer: (c).

4. Let $\Delta^{2}$ be the discriminant and $\alpha, \beta$ be the roots of the equation $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0$. Then $2 \mathrm{a} \alpha+\Delta$ and $2 \mathrm{a} \beta-\Delta$ can be roots of the equation

(a). $\mathrm{x}^{2}+2 \mathrm{bx}+\mathrm{b}^{2}=0$

(b). $\mathrm{x}^{2}-2 \mathrm{bx}+\mathrm{b}^{2}=0$

(c). $x^{2}+2 b x-3 b^{2}-16 a c=0$

(d). $\mathrm{x}^{2}+2 \mathrm{bx}-3 \mathrm{~b}^{2}+16 \mathrm{ac}=0$

Show Answer

Solution : $\alpha, \beta=\frac{-b \pm \sqrt{\Delta^{2}}}{2 a}$

$\therefore$ quadatic equation is

  • $\mathrm{x}^{2}+2 \mathrm{bx}-3 \mathrm{~b}^{2}+16 \mathrm{ac}=0$

$\therefore$ quadratic equation is

  • $\mathrm{x}^{2}+2 b \mathrm{bx}+\mathrm{b}^{2}=0$

Answer : (a and d)

5. The polynomial equation $\left(a x^{2}+b x+c\right)\left(a x^{2}-d x-c\right)=0, a c \neq 0$ has

(a). four real roots

(b). atleast two real roots

(c). atmost two real roots

(d). No real roots

Show Answer

Solution : $\mathrm{ac} \neq 0$

$\Rightarrow \quad \mathrm{ac}>0$ or $\mathrm{ac}<0$

Now $\mathrm{D} _{1}=\mathrm{b}^{2}-4 \mathrm{ac} \& \mathrm{D} _{2}=\mathrm{d}^{2}+4 \mathrm{ac}$

When ac $>0 \Rightarrow \mathrm{D} _{2}>0$ but $\mathrm{D} _{1}$ may positive or negative

When ac $<0 \Rightarrow \mathrm{D} _{1}>0$ but $\mathrm{D} _{2}$ may be positive or negative In either case the polynomial has atleast two real roots

Answer: (b).

6. If $\alpha, \beta$ are roots of $x^{2}-p(x+1)-q=0$, then the value of $\frac{\alpha^{2}+2 \alpha+1}{\alpha^{2}+2 \alpha+q}+\frac{\beta^{2}+2 \beta+1}{\beta^{2}+2 \beta+q}$ is

(a). 1

(b). 2

(c). 3

(d). None of these

Show Answer

Solution : $\mathrm{x}^{2}-\mathrm{px}-\mathrm{p}-\mathrm{q}=0$

$ \begin{aligned} & \Rightarrow \alpha+\beta=\mathrm{p} \\ & \text { and } \alpha \beta=-p-q \\ & \operatorname{Now}(\alpha+1)(\beta+1) \quad=(\alpha+\beta)+\alpha \beta+1 \\ & =\mathrm{p}-\mathrm{q}-\mathrm{p}+1=1-\mathrm{q} \\ & \text { Now } \frac{\alpha^{2}+2 \alpha+1}{\alpha^{2}+2 \alpha+q}+\frac{\beta^{2}+2 \beta+1}{\beta^{2}+2 \beta+q} \\ & =\frac{(\alpha+1)^{2}}{(\alpha+1)^{2}+q-1}+\frac{(\beta+1)^{2}}{(\beta+1)^{2}+q-1} \\ & =\frac{(\alpha+1)^{2}}{(\alpha+1)^{2}-(\alpha+1)(\beta+1)}+\frac{(\beta+1)^{2}}{(\beta+1)^{2}-(\alpha+1)(\beta+1)} \\ & =\frac{1}{1-\left(\frac{\beta+1}{\alpha+1}\right)}+\frac{1}{1-\frac{(\alpha+1)}{\beta+1}}=\frac{\alpha+1}{(\alpha+1)-(\beta+1)}+\frac{\beta+1}{\beta+1-(\alpha+1)} \\ & =\frac{\alpha+1}{\alpha-\beta}+\frac{\beta+}{\beta-\alpha}=\frac{\alpha+1-\beta-1}{\alpha-\beta} \\ & \Rightarrow \frac{\alpha-\beta}{\alpha-\beta}=1 \end{aligned} $

Answer: (a).

7. Let $\alpha, \beta, \gamma$ be the roots of the equation $x^{3}+4 x+1=0$, then $(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1}$ equals

(a). 2

(b). 3

(c). 4

(d). 5

Show Answer

Solution : $\alpha+\beta+\gamma=0, \alpha \beta+\beta \gamma+\gamma \alpha=4, \alpha \beta \gamma=-1$

$\therefore \frac{1}{\alpha+\beta}+\frac{1}{\beta+\gamma}+\frac{1}{\gamma+\alpha}=-\frac{1}{\gamma}-\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{\alpha \beta+\beta \gamma+\gamma \alpha}{\alpha \beta \gamma}=-\left(\frac{4}{-1}\right)=4$

Answer : (c).

Practice questions

1. The minimum value of $f(\mathrm{x})=\mathrm{x}^{2}+2 \mathrm{bx}+2 \mathrm{c}^{2}$ is greatest than the maximum value of $g(x)=-x^{2}-2 c x+b^{2}$, then ( $x$ being a real )

(a). $|c|>\frac{|b|}{\sqrt{3}}$

(b). $\frac{|\mathrm{c}|}{\sqrt{2}}>|\mathrm{b}|$

(c). $-1<\mathrm{c}<\sqrt{2} \mathrm{~b}$

(d). Non real value of b & c exist

Show Answer Answer: (b)

2. If $\mathrm{P}(\mathrm{x})$ is a polynomial of degree less than or equal to 2 and $\mathrm{S}$ is the set of all such polynomials so that $\mathrm{P}(1)=1, \mathrm{P}(0)=0$ and $\mathrm{P}^{1}(\mathrm{x})>0 \forall \mathrm{x} \in[0,1]$, then $\mathrm{S}=$ Here, $\mathrm{P}(\mathrm{x})=\mathrm{b} \mathrm{x}^{2}+\mathrm{ax}+\mathrm{c}$

(a). $\phi$

(b). $\left\{(1-\mathrm{a}) \mathrm{x}^{2}+\mathrm{ax}, 0<\mathrm{a}<2\right\}$

(c). $\left\{(1-\mathrm{a}) \mathrm{x}^{2}+\mathrm{ax}, \mathrm{a}>0\right\}$

(d). $\left\{(1-\mathrm{a}) \mathrm{x}^{2}+\mathrm{ax}, 0<\mathrm{a}<1\right\}$

Show Answer Answer: (b)

3. In the quadratic equation $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0$ if $\Delta=\mathrm{b}^{2}-4 \mathrm{ac}$ and $\alpha+\beta, \alpha^{2}+\beta^{2}$ and $\alpha^{3}+\beta^{3}$ are in G.P, where $\alpha, \beta$ are the roots of the equation, then

(a). $\Delta \neq 0$

(b). $\mathrm{b} _{\Delta}=0$

(c). $\mathrm{c} _{\Delta}=0$

(d). $\Delta=0$

Show Answer Answer: (c)

4. If $a, b, c$ are the sides of a triangle $A B C$ such that $x^{2}-2(a+b+c) x+3 \lambda(a b+b c+c a)=0$ has real roots, then

(a). $\lambda<\frac{4}{3}$

(b). $\lambda>\frac{5}{3}$

(c). $\lambda \in\left(\frac{4}{3}, \frac{5}{3}\right)$

(d). $ \lambda \in\left(\frac{1}{3}, \frac{5}{3}\right)$

Show Answer Answer: (a)

5. Let $\alpha \& \beta$ be the roots of $x^{2}-6 x-2=0$, with $\alpha>\beta$. If $a _{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$, then the value of $\frac{\mathrm{a} _{10}-2 \mathrm{a} _{8}}{2 \mathrm{a} _{9}}$ is

(a). 1

(b). 2

(c). 3

(d). 4

Show Answer Answer: (c)

6. If $x^{2}-10 a x-11 b=0$ have roots $c \& d \cdot x^{2}-10 c x-11 d=0$ have roots $a \& b$, then $a+b+c+d$ is

(a). 1210

(b). 1120

(c). 1200

(d). None of these

Show Answer Answer: (a)

7. If $\mathrm{t} _{\mathrm{n}}$ denotes the $\mathrm{n}^{\text {th }}$ term of an A.P. and $\mathrm{t} _{\mathrm{p}}=\frac{1}{\mathrm{q}}$ and $\mathrm{t} _{\mathrm{q}}=\frac{1}{\mathrm{p}}$, then which of the following is necessarily a root of the equation $(p+2 q-3 r) x^{2}+(q+2 r-3 p) x+(r+2 p-3 q)=0$ is

(a). $t _{p}$

(b). $\mathrm{t} _{\mathrm{q}}$

(c). $t _{\mathrm{pq}}$

(d). $t _{p+q}$

Show Answer Answer: (c)

8. The curve $y=(\lambda+1) x^{2}+2$ intersect the curve $y=\lambda x+3$ in exactly one point, if $\lambda$ equals

(a). $\{-2,2\}$

(b). $\{1\}$

(c). $\{-2\}$

(d). $\{2\}$

Show Answer Answer: (c)

9. Read the passage and answer the following questions.

Consider the equation $x^{4}+(1-2 k) x^{2}+k^{2}-1=0$ where $k$ is real. If $x^{2}$ is imaginary, or $x^{2}<0$, the equation has no real roots. If $x^{2}>0$, the equation has real roots.

(i) The equation has no real roots if $\mathrm{k} \in$

(a). $(-\infty-1)$

(b). $(-1,1)$

(c). $\left(1, \frac{5}{4}\right)$

(d). $\left(\frac{5}{4}, \infty\right)$

Show Answer Answer: (a, d)

(ii) The equation has only two real roots if $\mathrm{k} \in$

(a). $(-\infty-1)$

(b). $(0, 1)$

(c). $(1, 2)$

(d). $(-1, 1)$

Show Answer Answer: (d)

(iii) The equation has four real roots if $\mathrm{k} \in$

(a). $(-\infty, 0)$

(b). $(-1,1)$

(c). $\left(1, \frac{5}{4}\right)$

(d). $(1, \infty)$

Show Answer Answer: (c)

10. If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$, then the value of $\left|\begin{array}{ccc}1 & \cos (\beta-\alpha) & \cos \alpha \\ \cos (\alpha-\beta) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1\end{array}\right|$ is

(a). $\sin (\alpha+\beta)$

(b). $\sin \alpha \sin \beta$

(c). $1+\cos (\alpha+\beta)$

(d). None of these

Show Answer Answer: (d)

11. If $(1+k) \tan ^{2} x-4 \tan x-1+k=0$ has real roots, then

(a). $\mathrm{k}^{2} \leq 5$

(b). $\tan \left(\mathrm{x} _{1}+\mathrm{x} _{2}\right)=2$

(c). for $\mathrm{k}=2, \mathrm{x} _{1}=\frac{\pi}{4}$

(d). for $\mathrm{k}=1, \mathrm{x} _{1}=0$

Show Answer Answer: (a, b, c, d)

12. If $p, q \in\{1,2,3,4\}$, the number of equations of the form $\mathrm{px}^{2}+\mathrm{qx}+1=0$ having real roots is

(a). 15

(b). 9

(c). 7

(d). 8

Show Answer Answer: (c)

13. In $\triangle \mathrm{PQR} \angle \mathrm{R}=\frac{\pi}{2}$. If $\tan \frac{\mathrm{p}}{2} \& \tan \frac{\mathrm{Q}}{2}$ are the roots of the equation $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0(\mathrm{a} \neq 0)$ then

(a). $\mathrm{a}+\mathrm{b}=\mathrm{c}$

(b). $\mathrm{b}+\mathrm{c}=0$

(c). $a+\mathrm{c}=\mathrm{b}$

(d). $\mathrm{b}=\mathrm{c}$

Show Answer Answer: (a)


sathee Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! ЁЯМРЁЯУЪЁЯЪАЁЯОУ

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on

Please select your preferred language
рдХреГрдкрдпрд╛ рдЕрдкрдиреА рдкрд╕рдВрджреАрджрд╛ рднрд╛рд╖рд╛ рдЪреБрдиреЗрдВ