Question

I do not understand how to do this equation
x^2+14x+4=0

Answer 1

It's an ugly one, but you can use the quadratic formula for solving quadratic equations, assuming you have learned this (and you should have).

Answer 2

and and subtract 49 from it to complete the square

Answer 3

Yep. As a reminder, for \(ax^2+bx+c = 0\), the roots are given by the quadratic formula, which you should memorize by heart.\[ x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}\]

Answer 4

Just one silly question that has nothing to do with the question, how do you get the right side expression to be a fraction like that? When I try, x comes in with it :P

Answer 5

(that is how do you do that with the equation option)

Answer 6

x^2 + 14x + 4 = 0

x^2 + 14x = -4

x^2 + 14x + 7^2 = -4 + 7^2

(x+7)^2 = 45

x+7 = +- sqrt(45)

x = -7 +- sqrt(45)

Answer 7

using the latex

\frac{top}{bottom}

Answer 8

\[\frac{top}{bottom}\]

Answer 9

The derivation of the quadratic formula is rather sneaky. There are many places it's online -- I don't remember it 100%. It's too long to derive every time you want to get roots.

Answer 10

Thanks amistre64, I have been longing for that!

Answer 11

\[\text{\ [\frac{top}{bottom}\ ]}\]
no spaces of course

Answer 12

I see i have to divide by 2 with 14 to get 7

Answer 13

yes, divide the 14 in half, then square it

Answer 14

its called completeing the square

Answer 15

ok

Answer 16

thank you for helping me to understand this i need to practice more

Answer 17

practice does help ;)

Answer 18

completing the square is how you prove the quadratic formula

Answer 19

the answer to the problem is -7-3sqrt 5and -7+3sqrt5, how did they get this answer

Answer 20

\[ax^2+bx+c=0\]
\[x^2+\frac{b}{a}x+\frac{c}{a}=0\]
\[x^2+\frac{b}{a}x+(\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a}=0\]
\[(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)-\frac{b^2}{4a}+\frac{4c}{4a}=0\]
\[(x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{4c}{4a^2}\]
\[(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\]
\[x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\]
\[x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\]
\[x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\]
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 21

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[x=\frac{-14\pm\sqrt{14^2-4(1)(4)}}{2(1)}\]
\[x=\frac{-14\pm\sqrt{196-16}}{2}\]
\[x=\frac{-14\pm\sqrt{180}}{2}\]
\[x=\frac{-14\pm\sqrt{9*4*5}}{2}\]
\[x=\frac{-14\pm3*2\sqrt{5}}{2}\]
\[x=\frac{-7\pm3\sqrt{5}}{1}\]

Answer 22

ok I think i may understand now

Answer 23

amistre64 - use \left( and \right) as parenthesis when you type your LaTeX to make them taller. ;-)

Answer 24

ok