What two numbers has sum equal to 9/4 and the product equal to 3/4?

1 Answer
Jan 7, 2017

Answer:

#9/8+sqrt(33)/8" "# and #" "9/8-sqrt(33)/8#

Explanation:

Notice that given two numbers #a# and #b#:

#(x-a)(x-b) = x^2-(a+b)x+ab#

So the two numbers we want will be the zeros of the quadratic polynomial:

#f(x) = x^2-9/4x+3/4#

So we find:

#0 = 64f(x)#

#color(white)(0) = 64(x^2-9/4x+3/4)#

#color(white)(0) = 64x^2-144x+48#

#color(white)(0) = (8x)^2-2(8x)(9)+81-33#

#color(white)(0) = (8x-9)^2-(sqrt(33))^2#

#color(white)(0) = ((8x-9)-sqrt(33))((8x-9)+sqrt(33))#

#color(white)(0) = (8x-9-sqrt(33))(8x-9+sqrt(33))#

Hence:

#x = 1/8(9+-sqrt(33)) = 9/8+-sqrt(33)/8#

That is, the two numbers are:

#9/8+sqrt(33)/8" "# and #" "9/8-sqrt(33)/8#