two numbers that differ by 3 have a product of 88 . find the possible points of numbers
Answers
Answered by
6
Answer:
let one no be x
then other number= x+3
x(x+3)= 88
x^2 + 3x= 88
x^2+3x-88=0
now by splitting the middle term
x^2+11x-8x-88=0
x(x+11) -8(x+11)
(x-8)(x+11)
so x=8 and -11
now one no is 8
other no is x+3=8+3=11
also x=-11
so other no is =x+3=-11+3= -8
hope u like it;))
Answered by
4
Explanation:
let one number be (x ).
then, other number will be (x +3).
According to condition (A/C) –
(x) × ( x+3) = 88
x² +3x = 88
x² + 3x - 88 = 0
x² +(11x -8x) -88 =0 [ using splitting the middle term method]
x² + 11x -8x -88 =0
x ( x + 11). -8 (x +11) = 0
(x-8) + ( x + 11) = 0
x -8 = 0 and x + 11 = 0
x = 8. and. x = -11
so the possible numbers can be
(8,11. ) and ( -11,-8 ) .
hope u get it.
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