Question

find the zeros of xsquare + 8 by 3 x + 4 by 3

Answer 1

Answer:

X= -2/3,-2

Step-by-step explanation:

$3 {x}^{2} + 8x + 4 = 0 \\ 3 {x}^{2} + 6x + 2x + 4 = 0 \\ 3x(x + 2) + 2(x + 2) \\ (3x + 2)(x + 2) = 0 \\ x = - \frac{2}{3} \\ x = - 2 \\ ans.$

thanks!

Answer 2

Find the value:

Step-by-step explanation:

$\ Given \ value: \\\\x^2 \ + \frac{8}{3}x+\frac{4}{3} =0 \\\\\frac{3x^2+8x+4}{3} =0\\\\\ Equation: \\\\ \ 3x^2+8x+4= 0 \\\\\ solution:\\\\\rightarrow \ 3x^2+8x+4= 0 \\\\\rightarrow \ 3x^2+(6+2)x+4= 0 \\\\\rightarrow \ 3x^2+6x+2x+4= 0 \\\\\rightarrow \ 3x(x+2)+2(x+2)= 0 \\\\\rightarrow \ (x+2)(3x+2)= 0 \\\\\rightarrow \ (x+2)=0 \ \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ (3x+2)= 0 \\\\\rightarrow \ x=-2 \ \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ 3x=-2$

$\rightarrow \ x=-2 \ \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ x=\frac{-2}{3}$

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