HackerRank Algebra and Calculus Quiz- 1

Given the equation as  P^2- 4P - 21 = 0.

As a and b are its roots, so a + b = 4 and ab = -21.

Given new roots are \frac{a}{b}   and \frac{b}{a}

Hence sum of roots = \frac{a}{b} +\frac{b}{a} = \frac{(a^2 + b^2)}{ab}

So new equation = P^2 - \frac{-58}{21} P+1 = 0

=> 21 P^2 + 58P + 21 = 0

Given,

x+\frac{1}{x}=7

Squaring both sides,

(x+\frac{1}{x})^2=7^2

x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=49

x^2+\frac{1}{x^2}+2=49

x^2+\frac{1}{x^2}=47

5a+\frac{1}{3a}=5

On multiplying by \frac{3}{5}

=> 5a+\frac{1}{3a}=5\times \frac{3}{5}

On squaring both sides,

=> 9a^2+\frac{1}{25a^2}+2\times 3a\times \frac{1}{5a}=9

=> 9a^2+\frac{1}{25a^2}=9-\frac{6}{5}

=  \frac{39}{5}

Cost Price of 1 cow = Rs. p

Cost Price of 1 goat = Rs. q

3p + 8q = 47200... (i)

⇒ 8p + 3q = 100200... (ii)

By equation (i) \times 3 - (ii) \times 8 ,

9p + 24q – 64p – 24q = 141600 – 801600

=> 55p = 660000

p=\frac{660000}{55}

=Rs. 12000

a^3-b^3=(a-b)^3+3ab(a-b)

98=2^3+3ab(2)

98=8+6ab

6ab=98-8

ab=\frac{90}{6}

ab=15

\frac{a}{b}+\frac{b}{a}=2

\frac{a^2+b^2}{ab}=2

a^2+b^2=2ab

a^2+b^2-2ab=0

(a-b)^2 = 0

x+\frac{1}{x}=2

x^2+1=2x

x^2+1-2x=0

(x-1)^2=0 [/latex<br /> <br /> x-1=0<br /> <br /> x=1<br /> <br /> x^5+\frac{1}{x^5}

Substitute the value of x=1 in above equation,

1^5+\frac{1}{1^5}

=2

= m^3 + n^3 + 3mn

= m^3 + n^3 + 3mn(m+n) [since,m+n=1]

= (m+n)^3=1

(a+\frac{1}{a})^2=3

a+\frac{1}{a}=\sqrt{3}

Cubing both sides,

(a+\frac{1}{a})^3=(\sqrt{3})^3

a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=3\sqrt{3}

a^3+\frac{1}{a^3}+3\sqrt{3}-3\sqrt{3}=0

Therefore, a^6-\frac{1}{a^6}

(a^3+\frac{1}{a^3})(a^3-\frac{1}{a^3})=0

a + b + c = 1

ab + bc + ca = –1

(a + b + c)^2 = a^2+b^2+c^2+ 2(ab + bc + ca)

=> 1 =a^2+b^2+c^2+ 2 (–1)

=> a^2+b^2+c^2= 3

Therefore, a^3+b^3+c^3 = 3abc + 4

=> –3 + 4 = 1