Question

solve the quadratic equation 3x²-4x+20/3=0​

Answer 1

Thus, the roots of the equation are $\frac{2 - 4 \sqrt{i} }{ 3}, \frac{2 + 4 \sqrt{i} }{ 3}$.

Given:

The given equation is 3x²-4x+20/3=0.

To find:

The roots.

Solution:

The given equation is 3x²-4x+20/3=0.

Multiplying the equation by 3 we get-

$9 {x}^{2} - 12x + 20$

Thus, the roots of the equation are given as follows-

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

Here a is 9, b is -12 and c is 20.

So, we can write-

$x = \frac{ - ( - 12) + \sqrt{ {( - 12)}^{2} - 4 \times 9 \times 20 } }{2 \times 9} \\ x = \frac{ + 12 + \sqrt{144 - 720} }{18} \\ x = \frac{ + 12 + \sqrt{ - 576} }{18} \\ x = \frac{12 + 24 \sqrt{i} }{18} \\ x = \frac{2 + 4 \sqrt{i} }{ 3}$

Another root is given as-

$x = \frac{ - ( - 12) - \sqrt{ {( - 12)}^{2} - 4 \times 9 \times 20 } }{2 \times 9} \\ x = \frac{ + 12 - \sqrt{144 - 720} }{18} \\ x = \frac{ + 12 - \sqrt{ - 576} }{18} \\ x = \frac{12 - 24 \sqrt{i} }{18} \\ x = \frac{2 - 4 \sqrt{i} }{ 3}$

Thus, the roots of the equation are $\frac{2 - 4 \sqrt{i} }{ 3}, \frac{2 + 4 \sqrt{i} }{ 3}$.

#SPJ6

Answer 2

We need to solve the given quadratic equation.

Given quadratic equation is

$3 {x}^{2} - 4x + 20 \div 3 = 0 \\ 3(3 {x}^{2} - 4x + 20 \div 3) = 3 \times 0 \\ 9 {x}^{2} - 12x + 20 = 0$

We need to find the roots of this equation.

Formula to find the roots of this equation is

$- b + \sqrt{ {b}^{2} - 4ac} \div 2a \: and \: - b - \sqrt{ {b}^{2} - 4ac} \div 2a$

Here in this equation,

$a = 9 \\ b = - 12 \\ c = 20$

Roots are,

$- ( - 12) + \sqrt{ {12}^{2} - 4 \times 9 \times 20 } \div 2 \times 9 \\ 12 + \sqrt{144 - 720} \div 18 \\ 12 + \sqrt{ - 576 } \div 18 \\ 12 + \sqrt{ {i}^{2} \times 576} \div 18 \\ 12 + 24i \div 18 \\ 2 + 4i \div 3$

Similarly, another root is

$2 - 4i \div 3$

Therefore, roots of the given equation are

$2 + 4i \div 3 \: and \: 2 - 4i \div 3$

#SPJ2