Mathematics: What are the solutions of 3x^2 + 14x + 16 = 0? x = -8/3, –2 x = –2, -3/8 x = 3/8, 2 x = 8/3, 2

Treat them as different functionsf(x)=2x^2g(x)=-x^2-5x-1the intersection points are the solutionsthe x values are the solutionsanswer is 3...

What are the solutions of 3x^2 + 14x + 16 = 0? - id32141604820558, kitkat033157

MayFlowers

16.01.2022

The correct option is A) x = -8/3, –2.

Step-by-step explanation:

Consider the provided equation 3x^2 + 14x + 16 = 0

The above equation is in the form of ax^2 +bx + c = 0

The quadratic formula to find the root of the equation is:

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ \$

By comparing the above equation with the general equation we can conclude that:

a = 3, b = 14, and c = 16

Substitute the respective values in the above formula:

$x={\frac {-14\pm {\sqrt {14^{2}-4(3)(16)}}}{2(3)}}\ \$

$x={\frac {-14\pm {\sqrt {196-192}}}{6}}\ \$

$x={\frac {-14\pm {\sqrt {4}}}{6}}\ \$

$x={\frac {-14\pm2}{6}}\ \$

$x={\frac {-14+2}{6}}\ \ } or\ {\displaystyle x={\frac {-14-2}{6}}\ \$

$x={\frac {-12}{6}}\ \ } or\ {\displaystyle x={\frac {-16}{6}}\ \$

$x=-2}\ \ } or\ {\displaystyle x={\frac {-8}{3}}\ \$

Hence the solutions of the provided equation is $x=-2}\ \ } or\ {\displaystyle x={\frac {-8}{3}}\ \$ .

Thus, the correct option is A) x = -8/3, –2.

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