Question

How do i find the two real irrational solutions of this equation?
G(x)=x^2+14x+18

Answer 1

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Answer 2

ok

Answer 3

hellp please im so lost

Answer 4

@hartnn

Answer 5

do you know quadratic formula ?

Answer 6

Compare your quadratic function with \(G(x)= ax^2+bx+c=0\)
find \[a=...?\\b=...?\\c=...?\\\] \[ \\ \sqrt{b^2-4ac}=...?\]
then the two solutions of x are:
\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 7

i'm still super confused...sorry

Answer 8

could you first find a,b,c only?
\(x^2+14x+18\)
\(ax^2+bx+c\)

Answer 9

so a=1 b=14 and c=18?

Answer 10

yes!
b^2-4ac = ... ?

Answer 11

196-72?

Answer 12

right?

Answer 13

@hartnn

Answer 14

yes!

Answer 15

what next? :)

Answer 16

just plug the values in the formula given

Answer 17

so plug a=1 b=14 and c=18 into the fraction thing you showed me? sorry i'm botherin u :/

Answer 18

yes, no prob
and you already found b^2-4ac in it

Answer 19

so -4|sqrt sign 124 over 2?

Answer 20

\((-14 \pm \sqrt{124})/2 \)

Answer 21

and you can simplify that

Answer 22

........im sorry i'm not sure i can :/

Answer 23

124 = 4*31

Answer 24

(-7 \(\pm\sqrt{31}\))

Answer 25

how did you find the 7?

Answer 26

-14/2 = -7

Answer 27

right.sorry so the answer has to be 2 solutions but the sign b4 means it can either b positive or negative so thats two solutions right?

Answer 28

yes,
\( -7-\sqrt{31} \\ -7+\sqrt{31}\)

Answer 29

thanks so much!

Answer 30

welcome ^_^