Question

1. The sum of a number and its reciprocal is 17/4. find the number.
2. The sum of two numbers is 8 and the difference of their square is 16. find the numbers.
3. Find two consecutive natural numbers whose square have the sum 221.
4.There are three consecutive positive integers such that the sum of squares of first integer and product of second and third is 191. find those integer.

Answer 1

$1)$Let $x$ be the number and its reciprocal be $\frac{1}{x}$

sum of a number and its reciprocal$=x+\frac{1}{x}=\frac{17}{4}$

$\Rightarrow 4x^2-17x+4=0$

$\Rightarrow 4x^2-16x-x+4=0$

$\Rightarrow 4x(x-4)-(x-4)=0$

$\Rightarrow (x-4)(4x-1)=0$

$\Rightarrow x=4,\frac{1}{4}$

$2)$Let $x$ and $y$ be the numbers

sum of two numbers$=x+y=8$

difference of their square$={(x-y)}^2=16$

$\Rightarrow x-y=\pm 4$

$\Rightarrow x+y=8,x-y=4$ or $x+y=8,x-y=-4$

$\Rightarrow x+y+x-y=8+4$ or $x+y+x-y=8-4$

$\Rightarrow 2x=12$ or $2x=4$

$\Rightarrow x=6$ or $x=2$

Put $x=6$ in $x+y=8$ we get $y=8-x=8-6=2$

Put $x=2$ in $x+y=8$ we get $y=8-x=8-2=6$

$\therefore$ the numbers are $(6,2)$ and $(2,6)$

$3)$Let $x$ and $x+1$ be the two consecutive natural numbers.

Sum of the square of two natural numbers$=x^2+{(x+1)}^2=221$

$\Rightarrow x^2+x^2+2x+1-221=0$

$\Rightarrow 2x^2+2x-220=0$

$\Rightarrow x^2+x-110=0$

$\Rightarrow x^2+11x-10x-110=0$

$\Rightarrow x(x+11)-10(x+11)=0$

$\Rightarrow (x+11)(x-10)=0$

$\therefore x=10$ since $x=-11\ne N$

$4)$Let the three consecutive positive integers be $x,x+1,x+2$

Given:sum of squares of first integer and product of second and third$=191$

$\Rightarrow x^2+(x+1)(x+2)=191$

$\Rightarrow x^2+x^2+3x+2-191=0$

$\Rightarrow 2x^2+3x-189=0$

$\Rightarrow 2x^2+21x-18x-189=0$

$\Rightarrow x(2x+21)-9(2x+21)=0$

$\Rightarrow (2x+21)(x-9)=0$

$\Rightarrow x=-\frac{21}{2},9$

$\therefore x=9$ since $-\frac{21}{2}\notin Z$

The integers are $9,10,11$