Help with signs, The sum of two numbers is 8, and their product is 9, 75, what are they?

Question

So, the answer in the book states: 1, 5 and 6, 5.

But when I do the Algebra, I get:

$x + y = 8$ $x + y = 8$ ; $x \cdot y = 9, 75$ $x*y = 9, 75$

x = 8 - y

9, 75 = (8 - y ) y

$y^{2} - 8 y - 9, 75$ $y^2 -8y - 9, 75$

And then quadratic formula:

$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2}$ $(- b +- sqrt(b^2 - 4ac)) /2$

Which gives:

$\frac{- b \pm \sqrt{64 - 4 (- 9, 75) (1)}}{2}$ $(- b +- sqrt(64 - 4(-9, 75)(1) )) /2$

$\frac{- b \pm \sqrt{64 + 39}}{2}$ $(- b +- sqrt(64 +39)) /2$

$\frac{- b \pm \sqrt{103}}{2}$ $(- b +- sqrt(103)) /2$

The way to get 1, 5 and 6, 5 would be to have 64 - 39 in the parrnethesis, but I always get a cancelation of signs, either with - y^2 or -9, 5.

Help please.

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Answer 1 · 2018-05-06T18:02:44

$1.5$ $1.5$ and $6.5$ $6.5$

You have the problem essentially worked out so:
$x + y = 8$ $x+y=8$
$x y = 9.75$ $xy=9.75$

Substitute for $x$ $x$
$x = - y + 8$ $x=-y+8$
$(- y + 8) y = 9.75$ $(-y+8)y= 9.75$

This is where there was a minor error, $- y \cdot y$ $-y*y$ is $- y^{2}$ $-y^2$

$- y^{2} + 8 y - 9.75 = 0$ $-y^2+8y-9.75=0$

$y = \frac{- 8 \pm \sqrt{64 - 4 (- 1) (- 9.75)}}{2 (- 1)}$ $y=(-8+-sqrt(64-4(-1)(-9.75)))/(2(-1))$

$y = \frac{- 8 \pm \sqrt{25}}{- 2}$ $y=(-8+-sqrt(25))/(-2)$

$y = 4 \pm \frac{5}{2}$ $y= 4+-5/2$

$y = 6.5 or 1.5$ $y= 6.5 or 1.5$

So you can decide which number you want for $x$ $x$ and which you want for $y$ $y$ :

The numbers are:
$1.5$ $1.5$ and $6.5$ $6.5$