Quadratic Equations

Let $f(x)=x^2+2(a-3) x+9$.

If 6 lies between the roots of $f(x)=0$, then we must have the following conditions:

(i) Disc $>0$ and (ii) $f(6)<0$

[ $\because$ coefficient of $x^2$ is positive]

Now (i) Disc $>0$

$ \begin{array}{lc} \Rightarrow & 4(a-3)^2-36>0 \\ \Rightarrow & (a-3)^2-9>0 \\ \Rightarrow & a^2-6 a>0 \\ \Rightarrow & a<0 \text { or } a>6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ \text { and Eq. (ii) } & f(6)<0 \\ \Rightarrow & 36+12(a-3)+9<0 \\ \Rightarrow & 12 a+9<0 \\ \Rightarrow & a<-\frac{3}{4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i) and (ii), we get

                               $ a<-\frac{3}{4} $

i.e.                                $ a \in\left(-\infty, \frac{-3}{4}\right) $

For the quadratic equation, $a x^2+b x+c=0$, using the quadratic formula

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

Sum of roots $=-\frac{b}{a}$

Since, $a, b>0$, so $-\frac{b}{a}$ is negative, meaning the sum of the roots in negative.

Products of roots $=c / a$

Since, $a, c>0$, this means the product is positive we know that if the product of roots is positive and the sum of roots is negative, both roots must be negative.

To determine the nature of the roots of the equation $ x^2 + bx - c = 0 $ (where $ b, c > 0 $), we need to consider the sign of the product of the roots.

Key Points:

If the roots are of the same sign:

The product of the roots is positive.

If the roots are of opposite signs:

The product of the roots is negative.

Let's analyze the given equation $ x^2 + bx - c = 0 $:

The product of the roots for the quadratic equation $ ax^2 + bx + c = 0 $ is given by $ \frac{c}{a} $.

For our equation:

The product of the roots is $ \frac{-c}{1} = -c $.

Since $ c > 0 $, the product $-c$ is negative.

Conclusion:

The roots of the equation $ x^2 + bx - c = 0 $ must be of opposite signs, as the product of the roots is negative.

Let $ \log _{2}|x|=t \quad \ldots \text { (i) } $

$ \begin{array}{ll} \therefore & \sqrt{5-t}=3-t \\ \Rightarrow & 5-t=(3-t)^{2} \\ \Rightarrow & 5-t=9+t^{2}-6 t \end{array} $

$ \begin{array}{rrr} \Rightarrow & t^{2}-5 t+4 =0 \\ \Rightarrow & (t-4)(t-1) =0 \\ \Rightarrow & t =1 \text { or } 4 \text { [rejected] } \end{array} $

$ [\because \sqrt{5-4}=3-4,1 \neq-1] $

$ \begin{array}{lr} \Rightarrow & \log _{2}|x|=1 \\ \Rightarrow & |x|=2 \\ \Rightarrow & x= \pm 2 \end{array} $

$ \therefore $ Total 2 real solutions.

$\begin{aligned} & \text { Given, } 2 x^2-5 x+2=0 \\ & \Rightarrow \quad 2 x^2-4 x-x+2=0 \\ & \Rightarrow \quad 2 x(x-2)-1(x-2)=0 \\ & \Rightarrow \quad(2 x-1)(x-2)=0 \\ & \Rightarrow \quad x=\frac{1}{2}, 2 \quad \therefore \quad \alpha=\frac{1}{2} \\ & \text { Now, } 4 \alpha^2+4 \alpha=4\left(\frac{1}{2}\right)^2+4\left(\frac{1}{2}\right)=1+2=3 \neq 2 \\ & 4 \alpha^3-3 \alpha=4\left(\frac{1}{2}\right)^3-3\left(\frac{1}{2}\right)=\frac{1}{2}-\frac{3}{2}=-1 \neq 2 \\ & 4 \alpha^2-2 \alpha=4\left(\frac{1}{2}\right)^2-2\left(\frac{1}{2}\right)=1-1=0 \neq 2 \\ & 4 \alpha^3+3 \alpha=4\left(\frac{1}{2}\right)^3+3\left(\frac{1}{2}\right)=\frac{1}{2}+\frac{3}{2}=2 \end{aligned}$

Thus, other root be $4 \alpha^3+3 \alpha$.

Given $\alpha, \beta$ are the roots of $x^2-p x+r=0$

$\therefore \quad \alpha+\beta=p, \alpha \beta=r$

$\because \quad \frac{\alpha}{2}, 2 \beta$ are roots of $x^2-q x+r=0$

$\therefore \quad \frac{\alpha}{2}+2 \beta=q, \alpha \beta=r$

Now, $\alpha+\beta=p$ and $\alpha+4 \beta=2 q$

$\begin{aligned} \therefore \beta & =\frac{2 q-p}{3} \text { and } \alpha=p-\left(\frac{2 q-p}{3}\right) \\ & =\frac{2(2 p-q)}{3} \end{aligned}$

$\begin{aligned} \text { Hence, } r=\alpha \beta & =\left(\frac{2}{3}(2 p-q)\right)\left(\frac{2 q-p}{3}\right) \\ & =\frac{2}{9}(2 p-q)(2 q-p) \end{aligned}$

$4{x^2} + 2x - 1 = 0$

$\therefore$ $\alpha + \beta = - {1 \over 2},\alpha \beta = - {1 \over 4}$ ..... (i)

Also, $4{\alpha ^2} + 2\alpha - 1 = 0$ as $\alpha$ is a root and we have to prove that $\beta = 4{\alpha ^3} - 3\alpha $.

Now, $4{\alpha ^3} - 3\alpha = 4{\alpha ^2}\,.\,\alpha - 3\alpha $

$ = \alpha (1 - 2\alpha ) - 3\alpha = - 2\alpha - 2\alpha $

$ = - {1 \over 2}[4{\alpha ^2} + 4\alpha ) = - {1 \over 2}[1 - 2\alpha + 4\alpha ]$

$ = - {1 \over 2}(1 + 2\alpha ) = - {1 \over 2} - \alpha = \beta $

Now, $\alpha + \beta = - {1 \over 2}$ [from Eq. (i)]

Hence, $\beta = 4{\alpha ^3} - 3\alpha $.

Since, a + b = $-$p, ab = 1 ..... (i)

and c + d = $-$q, cd = 1

Now, (a $-$ c) (b $-$ c) and (a + d) (b + d) are the roots of x² + ax + $\beta$ = 0

(a $-$ c) (b $-$ c) (a + d) (b + d) = $\beta$

$\Rightarrow$ (ab $-$ ac $-$ bc + c²) (ab + ad + bd + d²) = $\beta$

$\Rightarrow$ {1 $-$ c (a + b) + c²} {1 + d(a + b) + d²} = $\beta$

$\Rightarrow$ (1 + pc + c²)(1 $-$ pd + d²) = $\beta$

$\Rightarrow$ 1 $-$ pd + d² + pc $-$ p²cd + pcd² + c² $-$ pc²d + c²d² = $\beta$

$\Rightarrow$ 1 $-$ pd + d² + pc $-$ p² + pd + c² $-$ pc + 1 = $\beta$ [$\because$ cd = 1]

$\Rightarrow$ 2 + d² + c² $-$ p² = $\beta$

$\Rightarrow$ 2cd + c² + d² $-$ p² = $\beta$

(c + d)² $-$ p² = $\beta$

q² $-$ p² = $\beta$ [$\because$ (c + d) = $-$ q]

Let $\alpha$, $\beta$ be the roots of x² $-$ abx $-$ a² = 0 where, $\alpha$ + $\beta$ = ab and $\alpha$$\beta$ = $-$a², which shows product of roots < 0, i.e. one root must be negative and the other must be positive. Hence, equation has one positive root and one negative root.

${4^{ - x + 0.5}} - {7.2^{ - x}} < 4$

Let ${2^{ - x}} = t$

The equation becomes

$2{t^2} - 7t - 4 < 0 \Rightarrow (t - 4)(2t + 1) < 0$

$ \Rightarrow 2(t - 4)\left\{ {t - \left( { - {1 \over 2}} \right)} \right\} < 0 \Rightarrow - {1 \over 2} < t < 4$

Since, $t = {2^{ - x}} > 0\forall x \in R$

$ \Rightarrow 0 < t < 4 \Rightarrow 0 < {2^{ - x}} < {2^2}$

As, 2^x is an increasing function,

$-$x < 2 or x > $-$2

Solution is $( - 2,\infty )$.

We have,

${x^2} - (k - 2)x + ({k^2} + 3k + 5) = 0$

x₁, x₂ are roots of equations

$\therefore$ ${x_1} + {x_2} = k - 2,{x_1}{x_2} = {k^2} + 3k + 5$

$x_1^2 + x_2^2 = {({x_1} + {x_2})^2} - 2{x_1}{x_2}$

$ = {(k - 2)^2} - 2({k^2} + 3k + 5)$

$ = {k^2} - 4k + 4 - 2{k^2} - 6k - 10$

$ = - ({k^2} + 10k + 6)$

$ = - [{(k + 5)^2} - 19]$

$ = 19 - {(k + 5)^2} \le 19$

$\therefore$ Maximum value of $x_1^2 + x_2^2 = 19$

Let f(x) = x¹⁰⁰

g(x) = x² $-$ 3x + 2

q(x) = quotient

r(x) = remainder = (ax + b)

$\therefore$ f(x) = g(x) q(x) + r(x)

x¹⁰⁰ = (x² $-$ 3x + 2) q(x) + ax + b

Put x = 1 $\to$ 1 = a + b ..... (i)

Put x = 2 $\Rightarrow$ 2¹⁰⁰ = 2a + b ..... (ii)

From Eqs. (i) and (ii), we get

a = 2¹⁰⁰ $-$ 1, b = 2 $-$ 2¹⁰⁰

Remainder = (2¹⁰⁰ $-$ 1)x + (2 $-$ 2¹⁰⁰)

= (2^{99 + 1} $-$ 1)x $-$ (2⁹⁹ $-$ 1)

$\therefore$ k = 99