x^2+√19x+15/4=0

Question

x^2+√19x+15/4=0

compphysgeek · Answer

you can use $x_{1,2} = - \frac{p}{2} \pm \sqrt{\frac{p^2}{4} - q}$ to solve for x when you've got a polynomial of the form $x^2 + px + q = 0$

anonymous · Answer

What is that formula called?

compphysgeek · Answer

we've just called it pq-formula .. dunno if it's got another name. you might also find it in a different form if the polynomial is of the form $ax^2 + bx + c = 0$

the solution then has the form $x_{1,2} = \frac{-a \pm \sqrt{a^2 - 4ac}}{2a}$ if I didn't make a mistake now. wrote it down from memory

compphysgeek · Answer

I did make a mistake since there is no b in my answer

compphysgeek · Answer

it should have been

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

anonymous · Answer

Well my homework says i have to solve the problem by completing the square, could you give me a break down on how to do it because my answer i got when i tried did not seem right

compphysgeek · Answer

ok .. completing the square means that you need to bring the equation into a form $$x^2 + px + (p/2)^2 + (q - p/2)= 0$$You should recognise a binomial equation here. $(a + b)^2 = a^2 + 2ab + b^2$

$(x + p/2)^2 = (p/2)^2 - q$ follows from that equation and if you solve the last one for x you'll get my first "pq-formula"

the p in your problem is $\sqrt{19}$. So we'll need to add $\sqrt{19}/2$ and subtract it to your problem. we're adding and subtracting the same number so that we dont change the value of the equation.

compphysgeek · Answer

in my previous answer the $\sqrt{19}/2$ that we add and subtract must be squared of course. so we're adding 19/4 and subtracting 19/4 as well.

$$x^2 + \sqrt{19}x + \frac{15}{4} + \frac{19}{4} - \frac{19}{4} = 0$$$$x^2 + \sqrt{19}x + \frac{19}{4} = \frac{19}{4} - \frac{15}{4} = 1$$$$(x + \sqrt{19}/2)^2 = 1$$$$x = 1 \pm \sqrt{19}/2$$

compphysgeek · Answer

and another mistake .. sorry

$$ x = \pm 1 - \sqrt{19}/2$$

anonymous · Answer

woah wait why are you dividing $$\sqrt{19}$$ by 2 and adding $$\sqrt{19}$$ divided by 4?

compphysgeek · Answer

another mistake? where's the edit button? 

the binomial equation $(x + b)^2 = x^2 + 2xb + b^2$ has a term 2bx. that 2b is $\sqrt{19}$ in your problem. that's why I add a term $(\sqrt{19}/2)^2\ = 19/4$ to the equation in your problem and subtract the same term again so that it doesn't change the value of the equation. it must still be 0

compphysgeek · Answer

In your equation, the term $b^2$ is not present. that's the one that I'm adding to complete the square.

compphysgeek · Answer

if we write it out again, your problem is $$x^2 + \sqrt{19}x = -\frac{15}{4}$$
to make the left hand side a square we're adding $(\frac{\sqrt{19}}{2})^2$ to both sides of the equation
$$ x^2 + \sqrt{19}x + \frac{19}{4} = \frac{19}{4} - \frac{15}{4}$$
we don't change the equation by adding the same term to both sides of the equation.
now the left hand side is a square and the right hand side is 1

compphysgeek · Answer

sorry if I'm confusing you

anonymous · Answer

okay i understand so far but the next part is where im getting messed up

compphysgeek · Answer

left hand side can now be written as $$x^2 + \sqrt{19}x + 19/4 = (x + \sqrt{19}/2)^2$$and we'll get $$(x + \sqrt{19}/2)^2 = 1$$
square root of both sides gives us $$x + \sqrt{19}/2 = \pm 1$$
just bring $\sqrt{19}/2$ to the other side and isolate x
$$x = \pm1 - \sqrt{19}/2$$

compphysgeek · Answer

we'll have two solutions, $x = -1 - \sqrt{19}/2$ and $x = 1 - \sqrt{19}/2$

anonymous · Answer

thank you so much i understand it now!