mathematics-class-11-unit-05-chapter-04-problem-solving-quadratic-equations-l-4

Quadratic equation lecture 4

Question

Solve for real solutions:

1.) $~ \left|x^2-3 x-4\right| =9-\left|x^2-1\right|$

2.) $~ |(x-4)(x+1)| =9-|(x+1)(x-1)|$

3.) $~ |x-4||x+1| =9-|x+1||x-1|$

Quadratic equation lecture 4

Solution

$\begin{aligned} & \text { i) Let } x<-1 \\ & {[-(x-4)][-(x+1)]=9-[(-(x+1))(-(x-1))]} \\ & (x-4)(x+1)=9-(x+1)(x-1) . \\ & 2 x^2-3 x-14=0 \\ & (2 x-7)(x+2)=0 \Rightarrow x=\frac{7} {2},-2 \\ & \therefore x=-2 \end{aligned}$

Quadratic equation lecture 4

Solution

ii) Let

$-1 \leq x<+1$

$-(x-4)(x+1)=9+(x+1)(x-1)$

$-x^2+3 x+4=9+x^2-1$

$2 x^2-3 x+4=0$

$D<0$ , no real solution

Quadratic equation lecture 4

Solution

iii) Let $1 \leq x<4$

$-(x-4)(x+1)=9-(x+1)(x-1)$

$3 x+4=10 \quad\Rightarrow x=2$

Quadratic equation lecture 4

Solution

iv) $Let\quad x=4$ .

The roots are $\frac{7} {2},-2$

Since x $\geqslant 4$ , there is no solution.

$\therefore$ The my val sections are $x= \pm 2$ .

Quadratic equation lecture 4

Question

Consider $4 x^3-3 x-p=0$ , where $-1 \leq p<1$ Show that there exsist exactly one solution on $[\frac{1}{2},1]$ . and find out.

Quadratic equation lecture 4

Solution

$Let f(x) =4 x^3-3 x-p$

$f\left(y_2\right) =4\left(\frac{1}{2}\right)^3-3 \cdot \frac{1}{2}-p$

$=-(1+p)$

$f(1) =1-p$

$f(1) \cdot f\left(\frac{1}{2}\right) =(p+1)(p-1)=p^2-1 \leq 0$

Quadratic equation lecture 4

Solution

$f(1 / 2)$ & $f(1)$ have Opp. signs

$\exists x_0 \in(1 / 2,1) \text { s.t } f\left(x_0\right)=0 \text {. }$

Uniqueness. $f^{\prime}(x)=12 x^2-3$

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Quadratic equation lecture 4

Solution

$=12(x-\frac{1}{2})(x+\frac{1}{2})$

On $[y, 1],\quad f^{\prime}(x)>0$

$f(x)$ is mmantically increaring.

$\therefore f(x)=0$ has exactly one sol’n on [ $\frac{1}{2},1$ ]

Quadratic equation lecture 4

Solution

To find the root:

$\text { Put } x=\cos \theta$

$4 \cos ^3 \theta-3 \cos \theta=p$

$\cos 3 \theta=p$

$\Rightarrow 3 \theta=\cos ^{-1}(p) \Rightarrow \theta=\frac{1}{3} \cos ^{-1}(p)$

$\therefore \cos \theta=\cos \left(\frac{1}{3} \cos ^{-1}(p)\right)$

$\therefore x=\cos \left(\frac{1}{3} \cos ^{-1}(p)\right)$

Quadratic equation lecture 4

Question

Suppose $\int_0^1\left(1+\cos ^8 x\right)\left(a x^2+b x+c\right) d x=\int_0^2\left(1+\cos ^8 x\right)\left(a x^2+b x+c\right) d x$ where $a, b, c \in \mathbb{R}$ . Show that $\exists$ a root for the equation.

$x^2+b x+c=0 \quad \text { on }(1,2)$

Quadratic equation lecture 4

Solution

$\text { Take } f(x)=\int_0^x\left(1+\cos ^8 t\right)\left(a t^2+b t+c\right) d t$

Given $f(1)=f(2)$ .

By Rolle’s theorem, $\exists c \in(1,2)$ s.t.

$f^{\prime}(c)=0 .$

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Quadratic equation lecture 4

Solution

(ie) $\left(1+\cos ^8 k\right)\left(a k^2+b k+c\right)=0$

$\Rightarrow a k^2+b k+c=0$ .

Ex. If $a+b+c=0$ , then $3 a x^2+2 b x+c=0$ has atleast one roort on $(0,1)$ . Apply Role’s theorem

Quadratic equation lecture 4

Question

Let $f(x)=a x^2+b x+c>0 \quad \forall x \in \mathbb{R}$ Set $g(x)=f(x)+f^{\prime}(x)+f^{\prime \prime}(x)$ . Then show that $g(x)>0 \quad \forall x \in \mathbb{R}$ .

Quadratic equation lecture 4

Solution

$\Rightarrow D_f<0 \quad$ & $\quad a>0$

$g(x)=a x^2+b x+c+2 a x+b+2 a$

$=a x^2+(b+2 a) x+2 a+b+c$

$D_g=(b+2 a)^2-4 a(2 a+b+c)$

$=\left(b^2-4 a c\right)-4 a^2<0$

$\therefore g(x)>0 \forall x \in \mathbb{R}$

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Quadratic equation lecture 4

Question

If $\alpha$ and $\beta$ real roots of $a x^2+b x+c=0$ where $\alpha<-1$ and $\beta>1$ , then show that $1+\frac{c}{a}+\left|\frac{b}{a}\right|<0$ .

Quadratic equation lecture 4

Solution

We have

$a f(1)<0 ; a f(1)<0$

$a(a-b+c)<0 ; a(a+b+c)<0$

Dividing by $a^2$ ,

$1-\frac{b}{a}+\frac{c}{a}<0 ; 1+\frac{b}{a}+\frac{c}{a}<0$

(ie) $1 \pm \frac{b}{a}+\frac{c}{a}<0$

(ie) $1+\left|\frac{b}{a}\right|+\frac{c}{a}<0$ .

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Quadratic equation lecture 4

Question

Solve for real solutions:

$\log _{2 x+5}\left(6 x^2+17 x+5\right)=4-\log _{3 x+1}\left(4 x^2+20 x+25\right)$

Quadratic equation lecture 4

Solution

$\log _{2 x+5}((2 x+5)(3 x+1))=4-\log _{3 x+1}(2 x+5)^2$

$\log^{2 x+5} _{2 x+5} +\log^{3 x+1} _{2 x+5} =4-2 \log^{2 x+5} _{3 x+1}$

$1+\log3 x+1{3 x+1} _{2 x+5} =4-2 \log^{2 x+5)} _{3 x+1}$

$\begin{array}{l|l} \text { Put } y=\log^{3 x+1} _{2 x+5} & \frac{1}{y}=\log^{2 x+5} _{3 x+1} \\\\ 1+y=4-\frac{2}{y} \\\\ y^2-3 y+2=0 \Rightarrow y=1,2 . \end{array}$

Quadratic equation lecture 4

Solution

Take $y=1$

$\log^{3 x+1} _{2 x+5} =1$

$\Rightarrow 2 x+5=3 x+1$

$\Rightarrow x=4$ .

Take $y=2$ .

$\log^{3 x+1} _{2 x+5} =2$

$\Rightarrow(2 x+5)^2=3 x+1$

Quadratic equation lecture 4

Solution

(ie) $4 x^2+17 x+24=0$

Here $D<0$ . Thus are no real solution.

$\therefore x=4$ is the only real solution.

Quadratic equation lecture 4

Question

Suppose $1+\log _5\left(x^2+1\right) \geqslant \log _5\left(a x^2+4 x+a\right) \quad \forall x \in \mathbb{R}$ Find the range of $a$

Quadratic equation lecture 4

Solution

$1 \geqslant \log _5 \frac{a x^2+4 x+a}{x^2+1}$

$\log _5 5 \geqslant \log \frac{a x^2+4 x+a}{x^2+1}$

$\Rightarrow 5 \geqslant \frac{a x^2+4 x+a}{x^2+1}$

Note that $a x^2+4 x+a>0$ & $x^2+1>0$

Quadratic equation lecture 4

Solution

Consider $a x^2+4 x+a$

$D=16-4 a^2<0 \text { and } a>0$

$a^2>4, a>0$

$\therefore a>2$

$a>2$ or $a<-2$

$\Rightarrow a x^2+4 x+a \leq 5\left(x^2+1\right)$

$(5-a) x^2-4 x+(5-a) \geqslant 0 \quad \forall x \in \mathbb{R}$

Quadratic equation lecture 4

Solution

This is the $\Leftrightarrow 16-4(5-a)^2 \leq 0$ & $5-a>0$

$(5-a)^2 \geqslant 4, a<5$

$\therefore 5-a \geqslant 2 \text { or } 5-a \leqslant-2 ; a<5 \qquad \biggl[5-a \geqslant 2 \text { or } 5-a \leqslant-2\biggl]$

$a \leq 3 \text { or } a \geqslant 7 ; a<5$

$\Rightarrow a \leq 3$ . The range of a is $(2,3)$ .

Quadratic equation lecture 4

Question

Solve:

$x y+y z=63$

$x z+y z=23, \quad x, y, z \in \mathbb{N}$

Quadratic equation lecture 4

Solution

$z(x+y)=23 \quad \text { Note, } x+y \geqslant 2$

Since 23 is prime.

$z=1$ & $x+y=23$ $\Rightarrow y=23-x$

$y(x+z)=63$

$(23-x)(x+1)=63$

$x^2-22 x+40=0$

$x=2,20$

Quadratic equation lecture 4

Solution

Take $x=2, \Rightarrow y=21$

Tate $x=20 \Rightarrow y=3$

$(2,21,1)$ & $(20,3,1)$ are the 2 solution.

Thank you