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1)Find the quadratic equation

3(3x
{}^(2)+ 1)=12x​
2)Find the quadratic equation

9y {}^(2) - 12y - 14 = 0

User Friedbunny
by
7.3k points

1 Answer

4 votes

Explanation:

"find" ?

the equations are already there.

do you mean solve in the sense of finding the zero points of the equations ?

if so,

the general solution of a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

3(3x² + 1) = 12x

9x² + 3 = 12 x

9x² - 12x + 3 = 0

a = 9

b = -12

c = 3

x = (12 ± sqrt(144 - 4×9×3))/18 = (12 ± sqrt(144-108))/18 =

= (12 ± sqrt(36))/18 = (12 ± 6)/18

x1 = (12 + 6) / 18 = 18/18 = 1

x2 = (12 - 6) / 18 = 6/18 = 1/3 = 0.3333333...

9y² - 12y - 14 = 0

same thing, we just use y instead of x.

a = 9

b = -12

c = -14

y = (12 ± sqrt(144 - 4×9×-14))/18 = (12 ± sqrt(144+504)/18 =

= (12 ± sqrt(648))/18 = (12 ± sqrt(324×2))/18 =

= (12 ± 18×sqrt(2))/18

y1 = (12 + 18×sqrt(2))/18 = 12/18 + sqrt(2) = 2/3 + sqrt(2) =

= 0.942809042...

y2 = (12 - 18×sqrt(2))/18 = 12/18 - sqrt(2) = 2/3 - sqrt(2) =

= -0.747546896...

User Burak SARICA
by
8.4k points

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