Question

Find the solutions to the equation below 30x^2-28x+6 =0

Answer 1

Option A and E are the correct options.

Explanation:

30x²-28x+6=0

2(15x²-14x+3) = 0

Or 15x²-14x+3 = 0

15x²-9x-5x+3 = 0

3x(x-3) - 1(5x-3) = 0

(3x-1)(5x-3) = 0

Now by the zero product rule

3x-1 = 0

3x = 1

x = 1/3

And (5x-3) = 0

5x = 3

x = 3/5

So two solutions are (1/3, 3/5).

Therefore options A and E are the correct options.

Answer 2

For this case, we have the following equation of the second degree:

`30x ^ 2-28x + 6 = 0`

If we divide between 2 on both sides of the equation, we will have:

`15x ^ 2-14x + 3 = 0`

Where:

`a = 15\\b = -14\\c = 3`

The solutions will come from:

`x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}`

Substituting:

`x = \frac {- (- 14) \pm \sqrt {(- 14) ^ 2-4 (15) (3)}} {2 (15)}\\x = \frac {14 \pm \sqrt {196-180}} {30}\\x = \frac {14 \pm \sqrt {16}} {30}\\x = \frac {14 \pm4} {30}`

So, we have:

`x_ {1} = \frac{14 + 4} {30} = \frac {18} {30} = \frac {18} {30} = \frac {3} {5}\\x_ {2} = \frac {14-4} {30} = \frac {10} {30} = \frac {1} {3}\\`

Answer:

`x_ {1} = \frac {3} {5}\\x_ {2} = \frac {1} {3}`

Option A

Option E