Question

What is wrong with the solution?
x(x+3)=9
x=9 or x+3=9
x=6

So, the roots are 9
and 6.

Answer 1

Ok, to solve by that rule the right side of the equation has to be zero. You would have to multiply it out, combine like terms, and then solve.

Answer 2

multiply 9 to ?

Answer 3

Multiply out the x(x+3).

x^2+3x=9

Subtract 9 from both sides

x^2+3x-9=0

Answer 4

ohh i get it now thank you :)

Answer 5

Since it doesn't factor you have to use the quadratic formula

Answer 6

so im going to factor x^2 + 3x - 9 = 0 ?

Answer 7

ohh idk how to use the quadratic formula :((((

Answer 8

Ok, let me switch to my computer so I can draw this better

Answer 9

So the quadratic formula is\[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a}\] and in the equation the coefficients of your problem are the a, b, and c.
\[1x^2+3x-9=0\]1 is your a
3 is your b
-9 is your c

Answer 10

Now, since it's a plus or minus sign (so you get both roots) you will have 2 equations
The first one\[\frac{ -1+\sqrt{(3^2)-(4)(1)(-9)} }{ (2)(1) }\] And the second one\[\frac{ -1-\sqrt{(3^)-(4)(1)(-9)} }{(2)(1) }\]

Answer 11

ok

Answer 12

whats next

Answer 13

If you solve those, you get your roots.

Answer 14

\[\frac{ -1+\sqrt{45} }{2 }=2.854101966\] is the first equation. I'm not sure whether you need it completely simplified or in radical form.

The second equation is\[\frac{ -1-\sqrt{45} }{ 2 }=-3.854101966\]

Answer 15

thank you so much! i already fan(ned) :)

Answer 16

Aww Thanks! Glad I could help:)

Answer 17

Great work @candyme!

Answer 18

uhhh is the second equation 3squared too?

Answer 19

Yes, I didn't even notice the typo