Answer: First Option
[tex]x_1=-\frac{2}{3}[/tex] and [tex]x_2=-3[/tex]
Step-by-step explanation:
We have the following quadratic equation:
[tex]3x^2 +11x+6=0[/tex]
To solve this equation use the quadratic formula
For an equation of the form:
[tex]ax ^ 2 + bx + c = 0[/tex]
The quadratic formula is:
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
Note that in this case: [tex]a=3,\ b=11,\ c=6[/tex]
Then:
[tex]x=\frac{-11\±\sqrt{11^2-4(3)(6)}}{2(3)}[/tex]
[tex]x=\frac{-11\±\sqrt{121-72}}{6}[/tex]
[tex]x=\frac{-11\±\sqrt{49}}{6}[/tex]
[tex]x=\frac{-11\±7}{6}[/tex]
[tex]x_1=\frac{-11+7}{6}[/tex]
[tex]x_1=-\frac{2}{3}[/tex]
[tex]x_2=\frac{-11-7}{6}[/tex]
[tex]x_2=\frac{-18}{6}[/tex]
[tex]x_2=-3[/tex]
The solutions are: [tex]x_1=-\frac{2}{3}[/tex] and [tex]x_2=-3[/tex]
Answer:
-[tex]\frac{2}{3}[/tex],-3
Step-by-step explanation:
Given equation,
[tex]3x^2+11x+6=0[/tex]
By the middle term splitting,
[tex]3x^2 + (9+2)x + 6 = 0[/tex]
[tex]3x^2 + 9x + 2x + 6 = 0[/tex]
[tex]3x(x+3)+2(x+3)=0[/tex]
[tex](3x+2)(x+3)=0[/tex]
By zero product product,
3x + 2 = 0 or x + 3 = 0
x = [tex]-\frac{2}{3}[/tex] or x = -3
which is the solution of the given equation.
Hence, FIRST option is correct.
