mathematics-class-11-unit-05-chapter-01-problem-solving-quadratic-equations-l-1

Problems on quadratic equation lecture - 1

$\text{Solution of quadratic equation}$

$y=a x^2+b x+c, \quad a, b, c \in R, \quad a \neq 0$

i) To Solve: $a x^2+b x+c=0 \longrightarrow$ (1)
ii) The exact shape of the parabola

$x^2+1=0$

$D: ~ = b^2-4 a c$

If D > 0 , 2 real distinct solutions

If D = 0, We have only solution (real)

If D < 0 , No real solution, but it has solutions of the form $\alpha+i \beta$ & $\alpha-i \beta$ .

Problems on quadratic equation lecture - 1

$\text{Graph of quadratic equation}$

D > 0 , a > 0

D > 0 , a < 0

D = 0 , a > 0

D = 0 , a < 0

D < 0 , a > 0

D < 0 , a < 0

alt text

Problems on quadratic equation lecture - 1

Notes

1) For $ax^2 + bx + c = 0$

The point of maxima or minima are attained at $x = -\frac{b}{2a}$

2) $\alpha$ & $\beta$ roots of $a x^2+b x+c=0$

$\alpha+\beta=-b/a$

$\alpha \beta=c/a$

alt text

Problems on quadratic equation lecture - 1

Problem

$x^2+x-3=0$

$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

If $\alpha$ & $\beta$ are roots of the given equation.

$\alpha^3-4 \beta^2+19=\text { ? }$

Problems on quadratic equation lecture - 1

Solution

Let $a=\alpha^3-4 \beta^2+19$

$b =\beta^3-4 \alpha^2+19$

$a+b =\alpha^3+\beta^3-4\left(\alpha^2+\beta^2\right)+38$

$=(\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]-4\left((\alpha+\beta)^2-2 \alpha \beta\right)+38$

( $\because \alpha+\beta=-1$ , $\alpha \beta=-3$ )

$=(-1)[1+9]-4(1+6)+38 =0$

Problems on quadratic equation lecture - 1

Solution

$a-b =\alpha^3-\beta^3+4\left(\alpha^2-\beta^2\right)$

$=(\alpha-\beta)\left[(\alpha+\beta)^2-\alpha \beta\right]+4(\alpha+\beta)(\alpha-\beta)$

$=(\alpha-\beta)[1+3]+(\alpha-\beta)[-4]$

$a-b =0$

$\Rightarrow 2 a =0 .$

$\text { Hence } a=0$

Problems on quadratic equation lecture - 1

Example

$(x-a)(x-a-b)=1; a, b \in \R$

Prove that there are 2 real roots $\alpha$ & $\beta$ such that $\alpha< a<\beta$ .

Problems on quadratic equation lecture - 1

Solution

Put $x-a=y$

$y(y-b)=1$

$y^2-b y-1=0$

$D=b^2+4>0$

Problems on quadratic equation lecture - 1

Solution

Thus are 2 distinct real roots, say $\alpha$ & $\beta$ :

$\alpha \beta=-1 \Rightarrow \alpha$ & $\beta$ have opp. signs

$\alpha<0<\beta$

$x=a+y$

$ \alpha

alt text

Problems on quadratic equation lecture - 1

Example

Let $m \geqslant-1$

$x^2+2(m-2) x+m^2-3 m+3=0$

Given that this has 2 distinct roots $\alpha$ & $\beta$ . If $\alpha^2+\beta^2=6$ , find $m$ .

Problems on quadratic equation lecture - 1

Solution

$D=4(m-2)^2-4\left(m^2-3 m+3\right)$

$=-4 m+4>0$

$\Rightarrow m<1$

$\therefore-1 \leq m<1$

Problems on quadratic equation lecture - 1

Solution

$\alpha+\beta =-2(m-2)$

$\alpha \beta =m^2-3 m+3$

$6=\alpha^2+\beta^2 =(\alpha+\beta)^2-2 \alpha \beta$ $=2 m^2-10 m+10$

We get, $~ ~ 2 m^2-10 m+4=0$

$m=\frac{5 \pm \sqrt{17}}{2}$

$\therefore m=\frac{5-\sqrt{17}}{2}$

Problems on quadratic equation lecture - 1

Example

$2 x^2-3 x-1=0 \leftarrow p \text { is a root}$

$x^2+3 x-2=0 \leftarrow q \text { is a root }$

Given $p q \neq 1$ . Find the value of $\frac{p q+p+1}{q}$

Problems on quadratic equation lecture - 1

Solution

$2 p^2-3 p-1=0 \longrightarrow ~ (1)$

$q^2+3 q-2=0 \longrightarrow ~ (2)$

$-q^2\left[-1 - \frac{3}{q}+\frac{2}{q^2}\right]=0$

$\Rightarrow 2 \cdot \frac{1}{q^2}-\frac{3}{q}-1=0$

$\frac{1}{q}$ is a root of the equation $2x^2-3x-1=0$

Problems on quadratic equation lecture - 1

Solution

$\therefore p+\frac{1}{q}=\frac{3}{2}$ and $\frac {p}{q}=-\frac{1}{2}.$

$\frac{p q+p+1}{q} =p+\frac{p}{q}+\frac{1}{q}$

$=3 / 2-1 / 2=1$

Problems on quadratic equation lecture - 1

Example

$\frac{1}{x^2+2 x}+\frac{1}{x^2+6 x+8}+\frac{1}{x^2+10 x+24}=\frac{1}{5}-\frac{1}{x^2+14 x+48}$

Problems on quadratic equation lecture - 1

Solution

$\frac{1}{x(x+2)}+\frac{1}{(x+4)(x+2)}+\frac{1}{(x+6)(x+4)}+\frac{1}{(x+8)(x+6)}=\frac{1}{5}$

$\frac{1}{2}\left[\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+8}\right]=\frac{1}{5}$

$\frac{1}{2}\left[\frac{1}{x}-\frac{1}{x+8}\right]=\frac{1}{5}$

$x^2+8 x-20=0 \Rightarrow x=2$ & $-10 .$

Problems on quadratic equation lecture - 1

Example

If $2 x^2-3 x+\frac{2-3 x}{x^2}=1,$ find the product of the real roots, where, $x \neq 0$

Problems on quadratic equation lecture - 1

Solution

$2\left(x^2+\frac{1}{x^2}\right)-3\left(x+\frac{1}{x}\right)=1$

$2 y^2-3 y-4=1 \quad \biggl[y=x+\frac{1}{x}\biggl]$

$2 y^2-3 y-5=0$

$\Rightarrow y=5 / 2,-1$

Problems on quadratic equation lecture - 1

Solution

$y =-1 \Rightarrow x+\frac{1}{x}+1=0$

$\text { (ie) } x^2+x+1=0$ which has no real roots.

$y=5 / 2 \Rightarrow x+\frac{1}{x}=5 / 2$

$2 x^2-5 x+2=0$ whose roots are $\frac{1}{2}$ & 2.

Product = 1

Problems on quadratic equation lecture - 1

Example

If $x^2+x(1-3 \alpha)+2-\alpha=0$

Find $\alpha$ so that this equation has real roots.

Problems on quadratic equation lecture - 1

Solution

To have $b^2-4 a c \geqslant 0$ ,

$(1-3 \alpha)^2-4(2-\alpha) \geqslant 0$

$9 \alpha^2-9 \alpha+7 \alpha-7 \geqslant 0$

$(\alpha-1)(9 \alpha+7) \geqslant 0 .$

This is possible if $\alpha \geq 1$ or if $\alpha \leq-7 / 9$

$\therefore \alpha \in(-\infty,-7 / 9] \cup[1, \infty)$

Problems on quadratic equation lecture - 1

Example

$a x^2+b x+c=0 \longleftarrow \alpha$ is a root

$-a x^2+b x+c=0 \longleftarrow \beta$ is a root.

Show: $~ \frac{a}{2} x^2+b x+c=0$ has a root between $\alpha ~ \& ~ \beta$ .

Problems on quadratic equation lecture - 1

Solution

Take $p(x)=\frac{a}{2} x^2+b x+c$

$p(\alpha)=\frac{a}{2} \alpha^2+b \alpha+c=\left[a \alpha^2+b \alpha+c\right]-\frac{a}{2} \alpha^2=-\frac{a}{2} \alpha^2$

$p(\beta)=\frac{a}{2} \beta^2+b \beta+c=\left[-a \beta^2+b \beta+c\right]+\frac{3 a}{2} \beta^2=\frac{3 a}{2} \beta^2$

$\therefore p(\alpha) p(\beta)$ have opposite Signs.

$p(x)=0$ for some $x \in(\alpha, \beta)$ .

Problems on quadratic equation lecture - 1

Example

$2016 x^2+2017 x+1=0$ .

$x=-1$ is a solution

By dividing the expression $2016 x^2+2017 x+1 ~$ by $~ x+1 ~$ , we get, $~ 2016 x+1$

So, another factor is $~ 2016 x+1$

$\Rightarrow ~ x = -1 ~$ & $~ x=-1 / 2016$

Problems on quadratic equation lecture - 1

Example

$|x|^2-5|x|+6=0 .$

$y^2-5 y+6=0 \qquad\biggl[Put|x|=y\biggl]$

$\Rightarrow y=3,2 .$

$|x|=3 \Rightarrow x= \pm 3$

$|x|=2 \Rightarrow x= \pm 2$

Problems on quadratic equation lecture - 1

Examples

Case (i) $~ |x-2|^2+|x-2|-2=0$ Verify that Roots are 3 & -1.

Case (ii) $~$ $\left|x^2+4 x+3\right|+2 x+5=0$ .

Problems on quadratic equation lecture - 1

Cases

$x^2+4 x+3>0$

$(x+3)(x+1)>0$

$x<-3 \text { or } x>-1 .$

Problems on quadratic equation lecture - 1

Cases

Given equation becomes $x^2+4 x+3+2 x+5=0$

(ie) $x^2+6 x+8=0$

$(x + 4) (x + 2) =0$

$\Rightarrow x = -4 ~$ & $~ -2$

$x = -4$ is the only solution.

Problems on quadratic equation lecture - 1

Cases

Case (ii)

$\rArr x^2+4 x+3<0 \Leftrightarrow x \in(-3,-1)$

$\rArr x^2+4 x+3-2x-5=0$

$\rArr x^2+2 x-2=0$

$\rArr x=\frac{-2 \pm \sqrt{4+8}}{2}=-1 \pm \sqrt{3}$

$\therefore x=-1-\sqrt{3}$ is the only solution

$-4,-1-\sqrt{3}$ .

Thank you