Question

The product of two alternative multiples of 4 is eight more than five times their sum. What are the two numbers?

Answer 1

Answer:

Step-by-step explanation:

Lt the no: be a and a+8

ATQ => a(a+8) = 8 + 5(a+a+8)

           a^2 + 8a = 8 + 10a + 40

           a^2 - 2a - 48 = 0

           a^2 + 6a - 8a - 48 = 0

           a(a+6) - 8(a+6) = 0

           a = 8

          a + 8 = 16

so, the numbers are 8 and 16

Hope for brainliest!!!

Answer 2

Answer:

8 and 16

Step-by-step explanation:

Let the two alternative multiples of 4 be x and x +8

Now we are given that The product of two alternative multiples of 4 is eight more than five times their sum.

The product of two alternative multiples of 4 = $x(x+8)$

The sum of two alternative multiples of 4 = $x+x+8$

A.T.Q

$x(x+8)=5(x+x+8)+8$

$x^2+8x=5(2x+8)+8$

$x^2+8x=10x+40+8$

$x^2+8x=10x+48$

$x^2=10x-8x+48$

$x^2=2x+48$

$x^2-2x-48=0$

$x^2-8x+6x-48=0$

$x(x-8)+6(x-8)=0$

$(x+6)(x-8)=0$

$x=-6,8$

- 6 is not a multiple of 4

So, Neglect -6

First multiple = 8

Second alternate multiple = x+8=8+8=16

Hence the two alternative multiples of 4 are 8 and 16.