Algebra Questions and Answers

We have log $a$ = $x$ log $b$
log $b$ = $y$ log $c$ and
loc $c$ = $z$ log $a$
 $therefore$ log $a$ log $b$ log $c$
= $xyz$ log $a$ log $b$
log $c$  $Rightarrow xyz = 1$

Given, $5x + 10y = 40$ i.e., $x + 2y = 8$ and $x + y = 7$.
After solving both the equations, we have $x = 6$ and $y = 1$

$= 1 + frac{x}{27} = sqrt{frac{784}{729}} = frac{28}{27} = 1 + frac{1}{27}$

Hence the  value of $x$ = 1

$text{R } = fleft(-3right) = 5left(-3right)^3 + 5left(-3right)^2 - 6(-3) + 9$

$= - 63$

$begin{aligned} & x^4 + y^4 - x^3y^2 - x^2y^3 + 16xy \ & [(x + y)^2 - 2xy]^2 - 4x^2y^2 + 16xy \ & (x + y)^4 - 4(x + y)^2 xy + 4x^2y^2 - 4x^2y^2 + 16xy \ & (x + y = 2 text{ given}) \ & (x + y)^4 = 2^4 = 2 times 2 times 2 times 2 = 16 end{aligned}$

$=alpha + beta = - frac{2b}{a}, alphabeta = frac{c}{a}$

$=sqrt{frac{alpha}{beta}} + sqrt{frac{beta}{alpha}} = frac{alpha + beta}{sqrt {alphabeta}} = frac{-frac{2b}{a}}{sqrt{frac{c}{a}}} = frac{-2b}{sqrt{ac}}$

$2^x = 2^{2y} = 2^{3z}$

$x = 2y =3z$

$(3z) biggl(frac{3}{2}z biggr)z = 288$

$text{ or } z^3 = frac{288 times 2}{9} = frac{576}{9} = 64$

$z = 4, x = 12, y = 6$

$frac{1}{2x} + frac{1}{4y} + frac{1}{8z} = frac{1}{24} + frac{1}{24} + frac{1}{24} = frac{3}{24} = 8$

$biggl(frac{a}{b}biggr)^{x-1} = biggl(frac{b}{a}biggr)^{x-3}$

$text{or }biggl(frac{a}{b}biggr)^{x-1} = biggl(frac{a}{b}biggr)^{-x-3}$

$therefore quad x-1 = -x + 3$

$2x = 4$

$x = 2$

$begin{aligned} 2^{x-7} &= 256 &= 2^8 therefore x-7 &= 8 text{or } x &= 15 end{aligned}$

By question, $x propto frac{1}{y^2} Rightarrow x = k quad frac{1}{y^2}$

[where $k$ is a constant]

$Rightarrow xy^2 = k$

Here $y = 2$ for $x = 1$

then $1 times 4 = k$

$Rightarrow k = 4$

Now, if  $y = 6 quad text{ then } quad x times 36 = 4$
$Rightarrow x = frac{1}{9}$

$f(2) = 3m$

$Rightarrow 16 - 8m + 8 - 10 + 8 = 3m$

$Rightarrow 22 - 8m = 3m Rightarrow m =2$

$text{Let } x^m = y^n = z^p = k$

$Rightarrow m log x = n log y = p log z = c$

$therefore log x = frac{c}{m}, log y = frac{c}{n}, log z = frac{c}{p}$

Now, given $frac{1}{m} + frac{1}{n} + frac{1}{p}$

$=frac{1}{c} (log x + log y + log z)$

$=frac{1}{c} log (xyz) = 0$

$[because log (xyz) = log 1 = 0]$

For $k = 4$, the equations are $x + 2y + 7 = 0$ and $2x + 4y + 14 = 0 = 2 (x + 2y + 7)$

$Bigg[because dfrac{a_1}{a_2} = dfrac{b_1}{b_2} = dfrac{c_1}{c_2}Bigg]$

Thus, the given equations have infinitely many solutions for $k = 4$

On the basis of quadratic equation, we solve the given equation for $x$, we get

$x = frac{-Big(2 - p^2Big) pm sqrt{Big(2 - p^2Big)^2 + 8p^2}}{2}$

$= frac{-2 + p^2 pm (2 + p^2)}{2}$

$= p^2, -2$

Let the rots of the given equation be M and $M text{ and }frac{1}{text{M}}$
 $therefore$ Product of the roots $= text{M } cdot frac{1}{text{M}} = frac{c}{a} = 1$

$therefore a = c$

Use Given equation with $alpha$ and $beta$ are roots, then
$alpha + beta = frac{-(-p)}{1} = p$

$alphabeta = frac{q}{1} = q$

Now the quadratic equation of roots $alphabeta + alpha + beta text{ and } alpha beta - alpha - beta$ is

$[x - (alphabeta + alpha + beta)] [x - (alphabeta - alpha - beta)] = 0$

$Rightarrow (x - q - p) (x - q + p ) = 0$

$Rightarrow x^2 - 2qx + q^2 - p^2 = 0$

Apply the formula,

$x = frac{-b pm sqrt{text{D}}}{2a}$

Given $4^{2x} = frac{1}{32} = 2^{-5}$

$Rightarrow 2^{4x} = 2^{-5}$

$Rightarrow 4x = -5 Rightarrow x = frac{-5}{4}$

Given, $10^{2y} = 25$

$Rightarrow (10^y)^2 = 25$

$Rightarrow 10^y = 5$

$therefore frac{1}{10^y} = frac{1}{5} Big[because frac{1}{10^y} = 10^{-y}Big]$

$because 10^{-y} = frac{1}{5}$

Given

$x = 2^frac{1}{3} + 2^frac{1}{3}$

on cubing both side, we get

$x^3 = 2 + 2^{-1} + 3 times 2^{frac{1}{3}} times 2^{-frac{1}{3}} (2^frac{1}{3} + 2^{-frac{1}{3}})$

$Rightarrow x^3 = 2 + frac{1}{2} + 3 times 2^frac{1}{3} times frac{1}{2^{1/3}} Big(2^frac{1}{3} + frac{1}{2^{1/3}}Big)$

$Rightarrow 2x^3 - 6x = 5$

Given

$a + b + c = 2s$

then the value of $(s - a)^2 + (s - b)^2 + (s - c)^2 + s^2$

$= s^2 + a^2 - 2as + s^2 + b^2 - 2bs + s^2 + c^2 - 2cs + s^2$

$begin{aligned} &=4s^2 - 2as - 2bs - 2cs + a^2 + b^2 + c^2 &=4s^2 - 2s(a + b + c) + a^2 + b^2 + c^2 &=4s^2 - 2s times 2s + a^2 + b^2 + c^2 &=a^2 + b^2 + c^2 end{aligned}$

Given,

$begin{aligned} &a^3 = 117 + b^3 & Rightarrow (3 + b)^3 = 17 + b^3 quadquad (because a = 3 + b) end{aligned}$

$begin{aligned} & Rightarrow 27 +b^3 + 9b (3 + b) = 117 + b^3 & Rightarrow b^2 + 3b - 10 = 0 & Rightarrow quadquad b = 2, -5, & because ;quadquad a = 3 + b & text{If } ;;quadquad b = 2, text{ then } a = 5 & text{If } ;;quadquad b = -5, text{ then } a = -2 & ;quadquadquad a + b = 2 + 5 & r - 2 - 5 text{ i.e. } pm 7 end{aligned}$

Given

$begin{aligned} a^{m + 5} &= 2^{2m + 10} &= 2^{2(m + 5)} therefore quad a^{m + 5} &= 4^{m + 5} therefore quad;;quad a &= 4 quadquad[because text{ Power is same}] end{aligned}$

Given, $(x^2 - 3x + 4)(x^2 + 2x + 5) = 0$

Let $alpha,beta$ be the roots of $x^2 - 3x + 4 = 0$

$therefore alpha + beta = 3$ and $alphabeta = 4$ (By using formula) and $gamma, delta$ be the roots of $x^2 + 2x + 5 = 0$

$therefore gamma + delta = -2$ and $gammadelta = 5$

$begin{aligned} & therefore alpha + beta + gamma + delta = 3 + (-2) = 1 = 5 text{(let) } & text{ and } alphabetagammadelta = 4 times 5 = 20 = text{P} text{(let)}end{aligned}$

$therefore$ The required equation is $x^2 - text{S}x + text{P} = 0$

$Rightarrow x^2 - x + 20 = 0$

Quadratic equation $x^2 - 2x + 1 < 0$

$Rightarrow (x - 1)^2 < 0,$which is impossible as the square of a real no. is never negative.

$therefore$ (D) is Answers

$begin{aligned} Given, x + y &= 1 therefore quad y &= 1 -x therefore quad xy &= x(1 - x) &= x - x^2 - frac{1}{4} + frac{1}{4} &=frac{1}{4} - Big(x^2 - x + frac{1}{4}Big) &= frac{1}{4} - Big(x - frac{1}{2}Big)^2 le frac{1}{4} &(because text{ square of a real no. } ge 0 ) therefore text{ Largest value } &= frac{1}{4} = 0.25 end{aligned}$

We know that $(x + y)^2 + (x - y)^2 = 2(x^2 + y^2)$

$therefore quad (x + y)^2 + 1^2 = 2(41)$

$therefore quad(x + y)^2 = 82-1$

$therefore quad x + y = sqrt{81} = pm 9$

Given, $sqrt{dfrac{x}{y}} + sqrt{dfrac{y}{x}} = dfrac{10}{3}$

$Rightarrow frac{big(sqrt{x}big)^2 + big(sqrt{y}big)^2}{sqrt{xy}} = frac{10}{3}$

$Rightarrow frac{x + y}{sqrt{xy}} = frac{10}{3}$

$Rightarrow frac{10}{sqrt xy} = frac{10}{3} quad [because x + y = 10]$

$therefore quad xy = 9$

$begin{aligned} a times b * c &= a times frac{sqrt{(b-c) + (b^2 - c^2)}}{2bc} \ \ &= 3 times frac{sqrt{(7 - 3) + (7^2 - 3^2)}}{2 times 7 times 3} \ \ &= 3 times frac{sqrt{4 + (49 - 9)}}{2 times 21} \ \ &= 3 times frac{sqrt{4 + 40}}{ 2 times 21} \ \ &= 3 times frac{2sqrt11}{2 times 21} = frac{sqrt11}{7} \ \ &=frac{sqrt11}{7} end{aligned}$