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find two numbers whose product is 12 such that one of the numbers is 4 less than 8 times the other number​

Sagot :

Answer:xy = 12

y = 8x - 4

 

x(8x-4) = 12

8x2 - 4x - 12 = 0

factor out a 4

4(2x2 - x - 3) = 0

2x2 - x - 3 = 0

 

we may be able to factor by grouping:

2(-3)= -6

Are there factors of -6 that add to -1?

(2)(-3) = -6 and 2-3=-1

Replace -x with 2x-3x and group

 

2x2 + 2x - 3x - 3 = 0

(2x2+2x) + (-3x-3) = 0

2x(x+1) -3(x+1) = 0

(2x-3)(x+1) = 0

 

Either 2x-3= 0    or    x+1=0

         2x=3               x = -1

          x = 3/2

 

we have 2 values of x

Find associated y values

 

x=3/2

y = 8x-4 = 8(3/2)-4 = 8

 

x = -1

y = 8x-4 = 8(-1)-4 = -12

 

your answers are:

(3/2, 8) and (-1, -12)

Step-by-step explanation: