Question

two numbers that differ by 3 have a product of 88 . find the possible points of numbers
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Answer 1

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;))

Answer 2

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.