Question

how mny real number solutions does the equation have? y=3x^2-5x-5 (1 point) the answer is two solutions but i need to show work. please help.

Answer 1

You can use the quadratic formula:
you have an equation of the form \(ax^2+bx+c\) where \(a=3, b=-5, c=-5\)

The formula is:
\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 2

so x= -5 + or - \[\sqrt{5^2}-4(3)(-5)\div2(3)\]

Answer 3

@kirbykirby

Answer 4

it's important to keep the square root over all of \(\sqrt{5^2-4(3)(-5)}\)

Answer 5

and -(-5) for the first term

Answer 6

i couldnt figure out how to do it.
but i think i messed up because i didnt make 5 negative

Answer 7

can u work the problem out with me

Answer 8

i got this so far
x=-5\[\pm \sqrt{85}\div6\]

Answer 9

@kirbykirby

Answer 10

\[ x=\frac{-(-5)\pm\sqrt{(-5)^2-4(3)(-5)}}{2(3)}=x=\frac{5\pm\sqrt{85}}{6}\]

Answer 11

so what next?
@kirbykirby

Answer 12

that's pretty much the solution, if you want though, you can write the roots separately:
\[ x=\frac{5}{6}+\frac{\sqrt{85}}{6} \text{and} \,\,\,\frac{5}{6}-\frac{\sqrt{85}}{6}\]

Answer 13

thx
@kirbykirby

Answer 14

but can you explain how that is 2 solutions
@kirbykirby

Answer 15

the \(\pm\) is just a way of saying "plus" or "minus" in a compact way rather than writing 2 different equations. If you prefer, you can write the quadratic as two equations (giving two solutions):

\[ x=\frac{-b+\sqrt{b^2-4ac}}{2a}\] and
\[ x=\frac{-b-\sqrt{b^2-4ac}}{2a}\]