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

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

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Answer 1 · 2017-01-07T01:49:55

$\frac{9}{8} + \frac{\sqrt{33}}{8}$ $9/8+sqrt(33)/8" "$ and $\frac{9}{8} - \frac{\sqrt{33}}{8}$ $" "9/8-sqrt(33)/8$

Explanation:

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

$(x - a) (x - b) = x^{2} - (a + b) x + a b$ $(x-a)(x-b) = x^2-(a+b)x+ab$

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

$f (x) = x^{2} - \frac{9}{4} x + \frac{3}{4}$ $f(x) = x^2-9/4x+3/4$

So we find:

$0 = 64 f (x)$ $0 = 64f(x)$

$0 = 64 (x^{2} - \frac{9}{4} x + \frac{3}{4})$ $color(white)(0) = 64(x^2-9/4x+3/4)$

$0 = 64 x^{2} - 144 x + 48$ $color(white)(0) = 64x^2-144x+48$

$0 = {(8 x)}^{2} - 2 (8 x) (9) + 81 - 33$ $color(white)(0) = (8x)^2-2(8x)(9)+81-33$

$0 = {(8 x - 9)}^{2} - {(\sqrt{33})}^{2}$ $color(white)(0) = (8x-9)^2-(sqrt(33))^2$

$0 = ((8 x - 9) - \sqrt{33}) ((8 x - 9) + \sqrt{33})$ $color(white)(0) = ((8x-9)-sqrt(33))((8x-9)+sqrt(33))$

$0 = (8 x - 9 - \sqrt{33}) (8 x - 9 + \sqrt{33})$ $color(white)(0) = (8x-9-sqrt(33))(8x-9+sqrt(33))$

Hence:

$x = \frac{1}{8} (9 \pm \sqrt{33}) = \frac{9}{8} \pm \frac{\sqrt{33}}{8}$ $x = 1/8(9+-sqrt(33)) = 9/8+-sqrt(33)/8$

That is, the two numbers are:

$\frac{9}{8} + \frac{\sqrt{33}}{8}$ $9/8+sqrt(33)/8" "$ and $\frac{9}{8} - \frac{\sqrt{33}}{8}$ $" "9/8-sqrt(33)/8$

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

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