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Mathematics 19 Online
OpenStudy (anonymous):

find the zeros of this function: g(x)= x^2 + 6x + 34

hartnn (hartnn):

to get zeros of g(x), equate g(x) = 0 and you get a quadratic equation of the form ax^2+bx+c=0 Compare your quadratic equation with \(ax^2+bx+c=0\) find \[a=...?\\b=...?\\c=...?\] then the two roots of x are: \(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

OpenStudy (anonymous):

a= x b= 6 c= 34

hartnn (hartnn):

a=1 b= 6, c=34 is correct :)

OpenStudy (anonymous):

what do i do after that?

hartnn (hartnn):

calculate \(b^2-4ac=... ?\)

OpenStudy (anonymous):

i got -100..

hartnn (hartnn):

b^2-4ac = 36 - 4*34 = -100 is correct, so, \(\sqrt {b^2-4ac}= \sqrt{-100} =10i\) just substitute this in the quadratic formula above.

OpenStudy (anonymous):

what do i substitute?

hartnn (hartnn):

\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\) just plug in values! yo know a,b, and b^2-4ac..

OpenStudy (anonymous):

i put all the values in but i cant solve it because i dont have my calculator

OpenStudy (whpalmer4):

You don't need a calculator for this!

hartnn (hartnn):

you don't need calc b=6, a=1 \(\huge{x_{1,2}=\frac{-6 \pm 10i}{2}}\)

OpenStudy (whpalmer4):

That can be simplified to \[-3\pm5i\]

OpenStudy (anonymous):

in the back of my book it says the answer is -3 (plus minus} 5i

OpenStudy (anonymous):

ahh okay thank you

OpenStudy (whpalmer4):

-6/2 = -3 10i/2 = 5i -10i/2 = -5i

OpenStudy (anonymous):

now since youre all here, how do you express each number in terms of i for 1/2 (squareroot) -16

OpenStudy (anonymous):

and (squareroot) -144

hartnn (hartnn):

note \(\sqrt{-n}= \sqrt{-1}\sqrt {n}= i \sqrt n\)

OpenStudy (whpalmer4):

remember that i^2 = -1 so \[\sqrt{-144} = \sqrt{-1*144}=i\sqrt{144}\]

OpenStudy (anonymous):

whpalmer4 : so what do i do after that?

OpenStudy (whpalmer4):

What is the square root of 144? Factor it: 144= 2*72 = 2*2*36 = 2*2*2*18 = 2*2*2*2*9=2*2*2*2*3*3 For every pair of identical factors, move one of them outside of the square root sign. \[\sqrt{144}=\sqrt{(2*2)*(2*2)*(3*3)}=2*2*3=12\]

OpenStudy (whpalmer4):

So \[\sqrt{-144}=i\sqrt{144}=12i\]

OpenStudy (anonymous):

yes i got that. what about1/2 \[\sqrt{-16}\]

OpenStudy (anonymous):

\[1/2\sqrt{-16}\]

hartnn (hartnn):

can't you find square root of 16 ?

OpenStudy (anonymous):

i got 2i

hartnn (hartnn):

yes, square root of 16 = 4 and (1/2) 4 = 2 and don't forget to have negative root, +2i and -2i

OpenStudy (anonymous):

okay thanks

hartnn (hartnn):

welcome ^_^

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