Question

the product of two consecutive positive integers is 306.find the integets.with full process

Answer 1

here you go ☆☆

▪let no. be X
▪so other no. =X + 1

▪product of two no. = 306

•(x)(X+1)=306

•x^2+ x = 306

•${x}^{2} + x - 306 = 0$

▪by using formula:-

•${b}^{2} - 4ac$

•${1}^{2} - 4(1)( - 306)$

•$1 + 1224$

▪1225

•$\frac{ - b( + - ) \sqrt{ {b}^{2} - 4ac } }{2a}$

•${1} (+ -) \sqrt{1225} \div 2$

•$\frac{ - 1( + - )35}{2}$

•$\frac{ - 1 + 35}{2} .. \frac{ - 1 - 35}{2}$

•$\frac{34}{2} .. \frac{ - 36}{2}$

▪two zeroes are
▪17
▪ -18

■-18 is neglected...as no. are positive.....

▪so no. are
•17
• 18

hope it helps you.....✌✌

Answer 2

let he base be x
x 1+x+2 =306
2x+3=306
2x=306-3
2x=303
x = 303/2
x+1. 303/2+1
x+2= 303/2 +2