Question

How do I solve 20-2.8x=5*sqrt(x)

Answer 1

On the right side of the equation, is it 5 times the square root of x or 5 - the square root of x?

Answer 2

times 5

\[20-2.8x=5\sqrt{x}\]

Answer 3

First, you can square both sides, getting 25x = \[(20-2.8x)^{2}\]

Answer 4

why did it become 25x?

Answer 5

Then, if you expand the right side, you would get \[25x = 7.84x ^{2} -112x + 400\]

Answer 6

shouldnt it stay 5x on the right side?

Answer 7

Because 5*sqrt(x) squared will square both terms. 5 squared is 25 and sqrt(x) squared is just x so it is 25x

Answer 8

But you take the square root away to the other side so only the other side will be squared right? :P

And btw the abc formula gives me a negative discriminant, whihc I cannot use in the -b+/-\[(-b+/-\sqrt{b ^{2}-4ac})/2a\]

Answer 9

The answer is about 3.7

Answer 10

That formula only works if the equation is equal to 0, which I am getting to...

Answer 11

Continuing, I will get all the terms on one side by subtracting 7.84 x^2 - 112x + 400 from both sides getting \[-7.84x ^{2} + 137 - 400 = 0\]

Answer 12

That is supposed to say 137x. Sorry for the mistake...

Answer 13

Both x's will become negative if I use that one into the abc formula

Answer 14

I am not sure what you are saying. Both x's will actually be positive. In fact, using the quadratic equation, which I am not going to type out here because you already know it, you will get the two answers x = 13.769 and x = 3.70544

Answer 15

@Krokodzilla notice that b = -137 and thus a really high number thus yielding a positive radical

Answer 16

a=-7,84
b=137
c=-400

Answer 17

\(\bf 20-2.8x=5\sqrt{x}\implies (20-2.8x)^2=(5\sqrt{x})^2
\\ \quad \\
20^2-2(20)(2.8x)+2.8^2=25x\implies 400-112x+7.84x^2=25x
\\ \quad \\

400-137x+7.84x^2=0\qquad
\textit{quadratic formula}\\
y={\color{blue}{ 7.84}}x^2{\color{red}{ -137}}x{\color{green}{ +400}}
\qquad \qquad
x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)

Answer 18

Yes, that is another way of looking at it. It is the same as my equation but multiplied by -1, which still gives the same answer. Excellent visual representation @jdoe0001

Answer 19

Ooh I see it now, i didnt squared the right side. Thank you both a lot! I really aprieciate it. :)

Answer 20

yw