Question

1/3x² + x+5/3x² = 1/9HAHAHAHA......​

Answer 1

Answer:

Now, 3x 2+x+5⩾0

This is because b² −4ac=1−4×5×3

=−59 (roots are imaginary)

3x²+x+5=(x−3)² =x²+9−6x and x−3⩾0

2x²+7x−4=0

2x²+8x−x−4=0

2x(x+4)−1(x+4)=0

(2x−1)(x+4)=0

x= 1/2 and−4 but x≥3

∴ No solution.

lol hehehe

Answer 2

Answer:

$\huge \bf༆ Answer ༄$

Step-by-step explanation:

Let's solve for x ~

• $\sf \dfrac{1}{3} {x}^{2} + x + \dfrac{5}{3} {x}^{2} = \dfrac{1}{9}$

• $\sf \dfrac{(1 + 5)}{3} {x}^{2} + x = \dfrac{1}{9}$

• $\sf \dfrac{6}{3} {x}^{2} + x - \dfrac{1}{9} = 0$

• $\sf2 {x}^{2} + x - \dfrac{1}{9} = 0$

Multiply each term with 9 on both sides of the equation

• $\sf(2 {x}^{2} \times 9) + (x \times9 ) - ( \dfrac{1}{9} \times 9 ) = 0 \times 9$

• $\sf18 {x}^{2} + 9x - 1 = 0$

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

$\sf \dfrac{ - {b}^{} \pm \sqrt{b {}^{2} - 4ac} }{2a}$

where,

• b is Coefficient of x = 9

• a is Coefficient of x² = 18

• c is the constant term = -1

Let's find the roots ~

• $\sf \dfrac{ - 9 \pm \sqrt{(9 {}^{2}) - (4 \times 18 \times - 1) } }{2 \times 18}$

• $\sf \dfrac{ - 9 \pm \sqrt{81- ( - 72) } }{36}$

• $\sf \dfrac{ - 9 \pm \sqrt{81 + 72} }{36}$

• $\sf \dfrac{ - 9 \pm \sqrt{153} }{36}$

• $\sf \dfrac{ - 9 \pm 3\sqrt{17} }{36}$

• $\sf \dfrac{3( - 3 \pm \sqrt{17} )}{36}$

• $\sf \dfrac{ - 3 \pm \sqrt{17} }{12}$

Therefore the value if x are ~

• $\sf \dfrac{ - 3 + \sqrt{17} }{12} \: \: and \: \: \sf \dfrac{ - 3 - \sqrt{17} }{12}$

I hope it helps ~