Algebra Questions Answers

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

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

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

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

$therefore quad xy = 9$

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