Quadratic Equation and Inequalities
Let $A = \{ x \in R:[x + 3] + [x + 4] \le 3\} ,$
$B = \left\{ {x \in R:{3^x}{{\left( {\sum\limits_{r = 1}^\infty {{3 \over {{{10}^r}}}} } \right)}^{x - 3}} < {3^{ - 3x}}} \right\},$ where [t] denotes greatest integer function. Then,
The sum of all the roots of the equation $\left|x^{2}-8 x+15\right|-2 x+7=0$ is :
The number of integral values of k, for which one root of the equation $2x^2-8x+k=0$ lies in the interval (1, 2) and its other root lies in the interval (2, 3), is :
Let $S = \left\{ {x:x \in \mathbb{R}\,\mathrm{and}\,{{(\sqrt 3 + \sqrt 2 )}^{{x^2} - 4}} + {{(\sqrt 3 - \sqrt 2 )}^{{x^2} - 4}} = 10} \right\}$. Then $n(S)$ is equal to
The number of real roots of the equation $\sqrt{x^{2}-4 x+3}+\sqrt{x^{2}-9}=\sqrt{4 x^{2}-14 x+6}$, is :
Let $\lambda \ne 0$ be a real number. Let $\alpha,\beta$ be the roots of the equation $14{x^2} - 31x + 3\lambda = 0$ and $\alpha,\gamma$ be the roots of the equation $35{x^2} - 53x + 4\lambda = 0$. Then ${{3\alpha } \over \beta }$ and ${{4\alpha } \over \gamma }$ are the roots of the equation
The number of real solutions of the equation $3\left( {{x^2} + {1 \over {{x^2}}}} \right) - 2\left( {x + {1 \over x}} \right) + 5 = 0$, is
The equation ${x^2} - 4x + [x] + 3 = x[x]$, where $[x]$ denotes the greatest integer function, has :
Let $[\alpha]$ denote the greatest integer $\leq \alpha$. Then $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots+[\sqrt{120}]$ is equal to __________
Explanation:
The number of points, where the curve $f(x)=\mathrm{e}^{8 x}-\mathrm{e}^{6 x}-3 \mathrm{e}^{4 x}-\mathrm{e}^{2 x}+1, x \in \mathbb{R}$ cuts $x$-axis, is equal to _________.
Explanation:
$f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1$ intersects the x-axis
where $f(x) = 0$. Setting $f(x)$ equal to zero gives us :
$e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0.$
Let $t = e^{2x}$. The equation now becomes :
$t^4 - t^3 - 3t^2 - t + 1 = 0.$
Dividing by $t^2$ and rearranging the equation gives :
$t^2 - t - 3 - \frac{1}{t} + \frac{1}{t^2} = 0.$
If we let $y = t + \frac{1}{t}$, we get :
$y^2 - y - 5 = 0.$
This quadratic equation in y can be solved using the quadratic formula to give two roots :
$y = \frac{1 \pm \sqrt{21}}{2}.$
Since $y = t + \frac{1}{t}$, and $t > 0$ (as $t = e^{2x}$), we must choose the root where $y > 2$. Thus, we take $y = \frac{1 + \sqrt{21}}{2}$.
So, we have :
$t + \frac{1}{t} = \frac{1 + \sqrt{21}}{2}.$
Solving for $t$ gives us :
$t^2 - yt + 1 = 0,$
or
$t = \frac{y \pm \sqrt{y^2 - 4}}{2}.$
Substituting $y = \frac{1 + \sqrt{21}}{2}$ into the formula gives :
$t = \frac{\frac{1 + \sqrt{21}}{2} \pm \sqrt{\left(\frac{1 + \sqrt{21}}{2}\right)^2 - 4}}{2}.$
This quadratic formula for $t$ yields two possible values (t = 2.37 and t = 0.42). Both of them are positive, thus both are acceptable values for $t = e^{2x}$.
Finally, to get $x$, take the natural log of both sides and divide by 2 :
$x = \frac{1}{2}\ln(t).$
This gives us two solutions for $x$, corresponding to the two positive solutions for $t$. So, the number of points where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$ intersects the x-axis is 2.
If $a$ and $b$ are the roots of the equation $x^{2}-7 x-1=0$, then the value of $\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$ is equal to _____________.
Explanation:
$ \begin{aligned} & x^2-7 x-1=0 \\\\ & \Rightarrow a^2-7 a-1=0 \Rightarrow a^2-1=7 a .........(i) \end{aligned} $
On squaring both sides, we get $a^4+1=51 a^2$
Similarly, $b^4+1=51 b^2$ ...........(ii)
$ \text { Now, } \frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}=\frac{a^{17}\left(a^4+1\right)+b^{17}\left(b^4+1\right)}{a^{19}+b^{19}} $
$ \begin{aligned} & =\frac{a^{17}\left(51 a^2\right)+b^{17}\left(51 b^2\right)}{a^{19}+b^{19}} \quad[\because \text { From Eq. (i) and (ii) }] \\\\ & =\frac{51\left[a^{19}+b^{19}\right]}{a^{19}+b^{19}}=51 \end{aligned} $
Let m and $\mathrm{n}$ be the numbers of real roots of the quadratic equations $x^{2}-12 x+[x]+31=0$ and $x^{2}-5|x+2|-4=0$ respectively, where $[x]$ denotes the greatest integer $\leq x$. Then $\mathrm{m}^{2}+\mathrm{mn}+\mathrm{n}^{2}$ is equal to __________.
Explanation:
$ \begin{aligned} & \Rightarrow\{x\}-x=x^2-12 x+31 \\\\ & \Rightarrow\{x\}=x^2-11 x+31 \end{aligned} $
So, $0 \leq x^2-11 x+31<1$
$ \begin{aligned} & \Rightarrow x^2-11 x+30 \leq 0 \\\\ & \Rightarrow(x-5)(x-6)<0 \\\\ & \Rightarrow x \in(5,6) \\\\ & \therefore[x]=5 \end{aligned} $
$ \begin{aligned} & \therefore x^2-12 x+5+31=0 \\\\ & \Rightarrow x^2-12 x+36=0 \\\\ & \Rightarrow(x-6)^2=0 \Rightarrow x=6 \end{aligned} $
Hence, $x \in \phi$
$ (\because x \in(5,6)) $
$ \therefore m=0 $
Another equation is $x^2-5[x+2]-4=0$
Case I : $x \geq-2$
$ x^2-5 x-14=0 \Rightarrow x=7,-2 $
Case II : $x<-2$
$ \begin{aligned} & x^2+5 x+6=0 \Rightarrow x=-3-2 \\\\ & \therefore x \in\{-3,-2,7\} \end{aligned} $
$ \therefore n=3 $
Hence, $m^2+m x+n^2=0+0+9=9$
have a common real root is $\frac{3}{\sqrt{2 \beta}}$ then $\beta$ is equal to ___________.
Explanation:
${x^2} - 5\alpha x + 1 = 0$ ..... (1)
${x^2} - \alpha x - 5 = 0$ ...... (2)
have a common root.
Subtracting (1) with (2) we'll get $x = {6 \over {4\alpha }}$
Substituting in (1)
${{36} \over {16{\alpha ^2}}} - {{30} \over 4} + 1 = 0$
$ \Rightarrow {\alpha ^2} = {9 \over {26}}$
$\alpha = {3 \over {\sqrt {2 \times 13} }}$
$\therefore$ $\beta = 13$
Let $\alpha_1,\alpha_2,....,\alpha_7$ be the roots of the equation ${x^7} + 3{x^5} - 13{x^3} - 15x = 0$ and $|{\alpha _1}| \ge |{\alpha _2}| \ge \,...\, \ge \,|{\alpha _7}|$. Then $\alpha_1\alpha_2-\alpha_3\alpha_4+\alpha_5\alpha_6$ is equal to _________.
Explanation:
${x^7} + 3{x^5} - 13{x^3} - 15x = 0$
$x({x^6} + 3{x^4} - 13{x^2} - 15) = 0$
$x = 0 = {\alpha _7}$
Let ${x^2} = t$
${t^3} + 3{t^2} - 13t - 15 = 0$
$(t + 1)(t + 5)(t - 3) = 0$
$t = {x^2} = - 1, - 5,3$
$x\, = \, \pm \,i, \pm \,\sqrt 5 i, \pm \,\sqrt 3 $
${\alpha _1},{\alpha _2} = \pm \,\sqrt 5 i,{\alpha _3},{\alpha _4} = \pm \,\sqrt 3 ,{\alpha _5},{\alpha _6} = \pm \,i$
${\alpha _1}{\alpha _2} - {\alpha _3}{\alpha _4} + {\alpha _5}{\alpha _6} = 5 + 3 + 1 = 9$
Let $\alpha \in\mathbb{R}$ and let $\alpha,\beta$ be the roots of the equation ${x^2} + {60^{{1 \over 4}}}x + a = 0$. If ${\alpha ^4} + {\beta ^4} = - 30$, then the product of all possible values of $a$ is ____________.
Explanation:
$ \therefore \alpha+\beta=-60^{\frac{1}{4}}, \alpha \beta=a $
Now $\alpha^{4}+\beta^{4}=-30$
$\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}-2 a^{2}=-30$
$\Rightarrow\left[(\alpha+\beta)^{2}-2 a\right]^{2}-2 a^{2}=-30$
$\Rightarrow\left(60^{\frac{1}{2}}-2 a\right)^{2}-2 a^{2}=-30$
$\Rightarrow 60+4 a^{2}-4 \cdot 60^{\frac{1}{2}} a-2 a^{2}+30=0$
$\Rightarrow 2 a^{2}-8 \sqrt{15} a+90=0$
Product of value of $a=45$
Let $\lambda \in \mathbb{R}$ and let the equation E be $|x{|^2} - 2|x| + |\lambda - 3| = 0$. Then the largest element in the set S = {$x+\lambda:x$ is an integer solution of E} is ______
Explanation:
$|\lambda-3| \leq 1$
$-1 \leq \lambda-3 \leq 1$
$2 \leq \lambda \leq 4$
$|x|=\frac{2 \pm \sqrt{4-4|\lambda-3|}}{2}$
$=1 \pm \sqrt{1-|\lambda-3|}$
$x_{\text {largest }}=1+1=2$, when $\lambda=3$
Largest element of $S=2+3=5$
If $\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$, then the maximum value of $\mathrm{a}$ is :
$
\text { Let } S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\} \text { and } $
$T=\left\{x \in \mathbb{Z}: x^{2}-7|x|+9 \leq 0\right\} \text {. }
$
Then the number of elements in $\mathrm{S} \cap \mathrm{T}$ is :
Let $\alpha$, $\beta$ be the roots of the equation $x^{2}-\sqrt{2} x+\sqrt{6}=0$ and $\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$ be the roots of the equation $x^{2}+a x+b=0$. Then the roots of the equation $x^{2}-(a+b-2) x+(a+b+2)=0$ are :
If $\alpha, \beta$ are the roots of the equation
$ x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0 $,
then the equation, whose roots are $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$, is :
The minimum value of the sum of the squares of the roots of $x^{2}+(3-a) x+1=2 a$ is:
If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^{4}+x^{3}+x^{2}+x+1=0$, then $\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}$ is equal to :
Let ${S_1} = \left\{ {x \in R - \{ 1,2\} :{{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0} \right\}$ and ${S_2} = \left\{ {x \in R:{3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0} \right\}$. Then, ${S_1} \cup {S_2}$ is equal to :
Let S be the set of all integral values of $\alpha$ for which the sum of squares of two real roots of the quadratic equation $3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$ is minimum. Then S :
Let $\alpha$ be a root of the equation 1 + x2 + x4 = 0. Then, the value of $\alpha$1011 + $\alpha$2022 $-$ $\alpha$3033 is equal to :
Let f(x) be a quadratic polynomial such that f($-$2) + f(3) = 0. If one of the roots of f(x) = 0 is $-$1, then the sum of the roots of f(x) = 0 is equal to :
The number of distinct real roots of x4 $-$ 4x + 1 = 0 is :
Let $A = \{ x \in R:|x + 1| < 2\} $ and $B = \{ x \in R:|x - 1| \ge 2\} $. Then which one of the following statements is NOT true?
Let a, b $\in$ R be such that the equation $a{x^2} - 2bx + 15 = 0$ has a repeated root $\alpha$. If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - 2bx + 21 = 0$, then ${\alpha ^2} + {\beta ^2}$ is equal to :
The sum of all the real roots of the equation
$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$ is
The number of distinct real roots of the equation
x7 $-$ 7x $-$ 2 = 0 is
If the sum of the squares of the reciprocals of the roots $\alpha$ and $\beta$ of
the equation 3x2 + $\lambda$x $-$ 1 = 0 is 15, then 6($\alpha$3 + $\beta$3)2 is equal to :
Let $\alpha, \beta(\alpha>\beta)$ be the roots of the quadratic equation $x^{2}-x-4=0 .$ If $P_{n}=\alpha^{n}-\beta^{n}$, $n \in \mathrm{N}$, then $\frac{P_{15} P_{16}-P_{14} P_{16}-P_{15}^{2}+P_{14} P_{15}}{P_{13} P_{14}}$ is equal to __________.
Explanation:
$\alpha$ and $\beta$ are the roots of the quadratic equation ${x^2} - x - 4 = 0$.
$\therefore$ $\alpha$ and $\beta$ are satisfy the given equation.
${\alpha ^2} - \alpha - 4 = 0$
$ \Rightarrow {\alpha ^{n + 1}} - {\alpha ^n} - 4{\alpha ^{n - 1}} = 0$ ...... (1)
and ${\beta ^2} - \beta - 4 = 0$
$ \Rightarrow {\beta ^{n + 1}} - {\beta ^n} - 4{\beta ^{n - 1}} = 0$ ...... (2)
Subtracting (2) from (1), we get,
$({\alpha ^{n + 1}} - {\beta ^{n + 1}}) - ({\alpha ^n} - {\beta ^n}) - 4({\alpha ^{n - 1}} - {\beta ^{n - 1}}) = 0$
$ \Rightarrow {P_{n + 1}} - {P_n} - 4{P_{n - 1}} = 0$
$ \Rightarrow {P_{n + 1}} = {P_n} + 4{P_{n - 1}}$
$ \Rightarrow {P_{n + 1}} - {P_n} = 4{P_{n - 1}}$
For $n = 14$, ${P_{15}} - {P_{14}} = 4{P_{13}}$
For $n = 15$, ${P_{16}} - {P_{15}} = 4{P_{14}}$
Now, ${{{P_{15}}{P_{16}} - {P_{14}}{P_{16}} - P_{15}^2 + {P_{14}}{P_{15}}} \over {{P_{13}}{P_{14}}}}$
$ = {{{P_{16}}({P_{15}} - {P_{14}}) - {P_{15}}({P_{15}} - {P_{14}})} \over {{P_{13}}{P_{14}}}}$
$ = {{({P_{15}} - {P_{14}})({P_{16}} - {P_{15}})} \over {{P_{13}}{P_{14}}}}$
$ = {{(4{P_{13}})(4{P_{14}})} \over {{P_{13}}{P_{14}}}}$
$ = 16$
The sum of all real values of $x$ for which $\frac{3 x^{2}-9 x+17}{x^{2}+3 x+10}=\frac{5 x^{2}-7 x+19}{3 x^{2}+5 x+12}$ is equal to __________.
Explanation:
${{3{x^2} - 9x + 17} \over {{x^2} + 3x + 10}} = {{5{x^2} - 7x + 19} \over {3{x^2} + 5x + 12}}$
$ \Rightarrow {{3{x^2} - 9x + 17} \over {5{x^2} - 7x + 19}} = {{{x^2} + 3x + 10} \over {3{x^2} + 5x + 12}}$
${{ - 2{x^2} - 2x - 2} \over {5{x^2} - 7x + 19}} = {{ - 2{x^2} - 2x - 2} \over {3{x^2} + 5x + 12}}$
Either ${x^2} + x + 1 = 0$ or No real roots
$ \Rightarrow 5{x^2} - 7x + 19 = 3{x^2} + 5x + 12$
$2{x^2} - 12x + 7 = 0$
sum of roots = 6
If for some $\mathrm{p}, \mathrm{q}, \mathrm{r} \in \mathbf{R}$, not all have same sign, one of the roots of the equation $\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right) x^{2}-2 \mathrm{q}(\mathrm{p}+\mathrm{r}) x+\mathrm{q}^{2}+\mathrm{r}^{2}=0$ is also a root of the equation $x^{2}+2 x-8=0$, then $\frac{\mathrm{q}^{2}+\mathrm{r}^{2}}{\mathrm{p}^{2}}$ is equal to ____________,
Explanation:
Let roots of
$\therefore$ $\alpha$ + $\beta$ > 0 and $\alpha$$\beta$ > 0
Also, it has a common root with ${x^2} + 2x - 8 = 0$
$\therefore$ The common root between above two equations is 4.
$ \Rightarrow 16({p^2} + {q^2}) - 8q(p + r) + {q^2} + {r^2} = 0$
$ \Rightarrow (16{p^2} - 8pq + {q^2}) + (16{q^2} - 8qr + {r^2}) = 0$
$ \Rightarrow {(4p - q)^2} + {(4q - r)^2} = 0$
$ \Rightarrow q = 4p$ and $r = 16p$
$\therefore$ ${{{q^2} + {r^2}} \over {{p^2}}} = {{16{p^2} + 256{p^2}} \over {{p^2}}} = 272$
The number of distinct real roots of the equation $x^{5}\left(x^{3}-x^{2}-x+1\right)+x\left(3 x^{3}-4 x^{2}-2 x+4\right)-1=0$ is ______________.
Explanation:
${x^8} - {x^7} - {x^6} + {x^5} + 3{x^4} - 4{x^3} - 2{x^2} + 4x - 1 = 0$
$ \Rightarrow {x^7}(x - 1) - {x^5}(x - 1) + 3{x^3}(x - 1) - x({x^2} - 1) + 2x(1 - x) + (x - 1) = 0$
$ \Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - x(x + 1) - 2x + 1) = 0$
$ \Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - {x^2} - 3x + 1) = 0$
$ \Rightarrow (x - 1)({x^5}({x^2} - 1) + 3x({x^2} - 1) - 1({x^2} - 1)) = 0$
$ \Rightarrow (x - 1)({x^2} - 1)({x^5} + 3x - 1) = 0$
$\therefore$ $x = \, \pm \,1$ are roots of above equation and ${x^5} + 3x - 1$ is a monotonic term hence vanishs at exactly one value of x other than 1 or $-$1.
$\therefore$ 3 real roots.
The number of real solutions of the equation ${e^{4x}} + 4{e^{3x}} - 58{e^{2x}} + 4{e^x} + 1 = 0$ is ___________.
Explanation:
Dividing by e2x
${e^{2x}} + 4{e^x} - 58 + 4{e^{ - x}} + {e^{ - 2x}} = 0$
$ \Rightarrow {({e^x} + {e^{ - x}})^2} + 4({e^x} + {e^{ - x}}) - 60 = 0$
Let ${e^x} + {e^{ - x}} = t \in [2,\infty )$
$ \Rightarrow {t^2} + 4t - 60 = 0$
$ \Rightarrow t = 6$ is only possible solution
${e^x} + {e^{ - x}} = 6 \Rightarrow {e^{2x}} - 6{e^x} + 1 = 0$
Let ${e^x} = p$,
${p^2} - 6p + 1 = 0$
$ \Rightarrow p = {{3 + \sqrt 5 } \over 2}$ or ${{3 - \sqrt 5 } \over 2}$
So $x = \ln \left( {{{3 + \sqrt 5 } \over 2}} \right)$ or $\ln \left( {{{3 - \sqrt 5 } \over 2}} \right)$
Let $\alpha$, $\beta$ be the roots of the equation ${x^2} - 4\lambda x + 5 = 0$ and $\alpha$, $\gamma$ be the roots of the equation ${x^2} - \left( {3\sqrt 2 + 2\sqrt 3 } \right)x + 7 + 3\lambda \sqrt 3 = 0$, $\lambda$ > 0. If $\beta + \gamma = 3\sqrt 2 $, then ${(\alpha + 2\beta + \gamma )^2}$ is equal to __________.
Explanation:
$\because$ $\alpha$, $\beta$ are roots of x2 $-$ 4$\lambda$x + 5 = 0
$\therefore$ $\alpha$ + $\beta$ = 4$\lambda$ and $\alpha$$\beta$ = 5
Also, $\alpha$, $\gamma$ are roots of
${x^2} - (3\sqrt 2 + 2\sqrt 3 )x + 7 + 3\sqrt 3 \lambda = 0,\,\lambda > 0$
$\therefore$ $\alpha + \gamma = 3\sqrt 2 + 2\sqrt 3 $, $\alpha \gamma = 7 + 3\sqrt 3 \lambda $
$\because$ $\alpha$ is common root
$\therefore$ ${\alpha ^2} - 4\lambda \,\,\alpha + 5 = 0$ ....... (i)
and ${\alpha ^2} - (3\sqrt 2 + 2\sqrt 3 )\alpha + 7 + 3\sqrt 3 \lambda = 0$ ...... (ii)
From (i) - (ii) : we get $\alpha = {{2 + 3\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$
$\because$ $\beta + \gamma = 3\sqrt 2 $
$\therefore$ $4\lambda + 3\sqrt 2 + 2\sqrt 3 - 2\alpha = 3\sqrt 2 $
$ \Rightarrow 3\sqrt 2 = 4\lambda + 3\sqrt 2 + 2\sqrt 3 - {{4 + 6\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$
$ \Rightarrow 8{\lambda ^2} + 3(\sqrt 3 + 2\sqrt 2 )\lambda - 4 - 3\sqrt 6 = 0$
$\therefore$ $\lambda = {{6\sqrt 2 - 3\sqrt 2 \pm \sqrt {9(11 - 4\sqrt 6 ) + 32(4 + 3\sqrt 6 )} } \over {16}}$
$\therefore$ $\lambda = \sqrt 2 $
$\therefore$ ${(\alpha + 2\beta + \gamma )^2} = {(\alpha + \beta + \beta + \gamma )^2}$
$ = {(4\sqrt 2 + 3\sqrt 2 )^2}$
$ = {(7\sqrt 2 )^2} = 98$
If the sum of all the roots of the equation
${e^{2x}} - 11{e^x} - 45{e^{ - x}} + {{81} \over 2} = 0$ is ${\log _e}p$, then p is equal to ____________.
Explanation:
$e^{2 x}-11 e^x-45 e^{-x}+\frac{81}{2}=0 $
$\Rightarrow 2 e^{3 x}-22 e^{2 x}-90+81 e^x=0 $
$\Rightarrow 2\left(e^x\right)^3-22\left(e^x\right)^2+81 e^x-90=0$
Let $ e^x=y$
$ \Rightarrow 2 y^3-22 y^2+81 y-90=0 $
Product of roots $\left(y_1, y_2, y_3\right)$
$ y_1 \cdot y_2 \cdot y_3=\frac{-(-90)}{2}=45 $
Let $x_1, x_2$, and $x_3$ be roots of given equation
$\Rightarrow e^{x_1} \cdot e^{x_2} \cdot e^{x_3} = 45 $
$\Rightarrow e^{x_1+x_2+x_3} =45$
$\Rightarrow x_1+x_2+x_3 =\log _e 45=\log _e p $
$\Rightarrow p = 45$
Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then ${\left( {{1 \over p} + {1 \over q}} \right)^{ - 2}}$ is equal to _________.
Explanation:
$\because$ $p + q = 3$ ...... (i)
and ${p^4} + {q^4} = 369$ ...... (ii)
${\{ {(p + q)^2} - 2pq\} ^2} - 2{p^2}{q^2} = 369$
or ${(9 - 2pq)^2} - 2{(pq)^2} = 369$
or ${(pq)^2} - 18pq - 144 = 0$
$\therefore$ $pq = - 6$ or 24
But $pq = 24$ is not possible
$\therefore$ $pq = - 6$
Hence, ${\left( {{1 \over p} + {1 \over q}} \right)^{ - 2}} = {\left( {{{pq} \over {p + q}}} \right)^2} = {( - 2)^2} = 4$
The sum of the cubes of all the roots of the equation
${x^4} - 3{x^3} - 2{x^2} + 3x + 1 = 0$ is _________.
Explanation:
${x^4} - 3{x^3} - {x^2} - {x^2} + 3x + 1 = 0$
$({x^2} - 1)({x^2} - 3x - 1) = 0$
Let the root of ${x^2} - 3x - 1 = 0$ be $\alpha$ and $\beta$ and other two roots of given equation are 1 and $-$1
So sum of cubes of roots
$ = {1^3} + {( - 1)^3} + {\alpha ^3} + {\beta ^3}$
$ = {(\alpha + \beta )^3} - 3\alpha \beta (\alpha + \beta )$
$ = {(3)^3} - 3( - 1)(3)$
$ = 36$
$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$, is :
equation x2 + (20)1/4x + (5)1/2 = 0. Then $\alpha$8 + $\beta$8 is equal to
then $\sum\limits_{n = 8}^{100} {\left[ {{{{{( - 1)}^n}n} \over 2}} \right]} $ is equal to :