1/3x² + x+5/3x² = 1/9

HAHAHAHA......​

Answers

Answer 1

Answer:

[tex]\huge \bf༆ Answer ༄[/tex]

Step-by-step explanation:

Let's solve for x ~

[tex] \sf \dfrac{1}{3} {x}^{2} + x + \dfrac{5}{3} {x}^{2} = \dfrac{1}{9} [/tex]

[tex] \sf \dfrac{(1 + 5)}{3} {x}^{2} + x = \dfrac{1}{9} [/tex]

[tex] \sf \dfrac{6}{3} {x}^{2} + x - \dfrac{1}{9} = 0[/tex]

[tex] \sf2 {x}^{2} + x - \dfrac{1}{9} = 0[/tex]

Multiply each term with 9 on both sides of the equation

[tex] \sf(2 {x}^{2} \times 9) + (x \times9 ) - ( \dfrac{1}{9} \times 9 ) = 0 \times 9[/tex]

[tex] \sf18 {x}^{2} + 9x - 1 = 0[/tex]

Now, let's use Quadratic formula to find its roots ~

[tex] \sf \dfrac{ - {b}^{} \pm \sqrt{b {}^{2} - 4ac} }{2a} [/tex]

where,

b is Coefficient of x = 9

a is Coefficient of x² = 18

c is the constant term = -1

Let's find the roots ~

[tex] \sf \dfrac{ - 9 \pm \sqrt{(9 {}^{2}) - (4 \times 18 \times - 1) } }{2 \times 18} [/tex]

[tex] \sf \dfrac{ - 9 \pm \sqrt{81- ( - 72) } }{36} [/tex]

[tex] \sf \dfrac{ - 9 \pm \sqrt{81 + 72} }{36} [/tex]

[tex] \sf \dfrac{ - 9 \pm \sqrt{153} }{36} [/tex]

[tex] \sf \dfrac{ - 9 \pm 3\sqrt{17} }{36} [/tex]

[tex] \sf \dfrac{3( - 3 \pm \sqrt{17} )}{36}[/tex]

[tex] \sf \dfrac{ - 3 \pm \sqrt{17} }{12} [/tex]

Therefore the value if x are ~

[tex] \sf \dfrac{ - 3 + \sqrt{17} }{12} \: \: and \: \: \sf \dfrac{ - 3 - \sqrt{17} }{12} [/tex]

I hope it helps ~


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