Question

Pllllllllllllllllllllleasee one guys i neeed ur help one

Answer 1

`{ \qquad\qquad\huge\underline{{\sf Answer}}}`

Let's solve ~

Calculate discriminant :

`\qquad \sf  \dashrightarrow \: 3 {x}^(2) + 6x - 1`

• a = 3
• b = 6
• c = 1

`\qquad \sf  \dashrightarrow \: discriminant = {b}^(2) - 4ac`

`\qquad \sf  \dashrightarrow \: d = (6) {}^(2) - (4 * 3 * 1)`

`\qquad \sf  \dashrightarrow \: d = 36 - 12`

`\qquad \sf  \dashrightarrow \: d = 24`

`\qquad \sf  \dashrightarrow \: \sqrt {d} = 2 √(6)`

Now, let's calculate it's roots ( x - intercepts )

`\qquad \sf  \dashrightarrow \: x = \cfrac{ - b \pm √(d) }{2a}`

`\qquad \sf  \dashrightarrow \: x = \cfrac{ - 6\pm 2 √(6) }{2 * 3}`

`\qquad \sf  \dashrightarrow \: x = \cfrac{ - 6\pm 2 √(6) }{6}`

So, the intercepts are :

`\qquad \sf  \dashrightarrow \: x = \cfrac{ - 6 - 2 √(6) }{6}`

and

`\qquad \sf  \dashrightarrow \: x = \cfrac{ - 6 + 2 √(6) }{6}`

Answer 2

`\left( ( -3 + 2√(3))/( 3), \ 0\right), \ \left(( -3 - 2√(3))/( 3), \ 0\right)`

Step-by-step explanation:

Given expression:

f(x) = 3x² + 6x - 1

• To find x intercepts, set f(x) = 0

Use quadratic formula:

`\sf x = ( -b \pm √(b^2 - 4ac))/(2a) \ where \ ax^2 + bx + c = 0`

Here after finding coefficients:

• a = 3, b = 6, c = -1

Applying formula:

`x = ( -6 \pm √(6^2 - 4(3)(-1)))/(2(3))`

`x = ( -6 \pm √(48))/(6)`

`x = ( -6 \pm 4√(3))/(6)`

`x = ( -6 \pm 4√(3))/(2 \cdot 3)`

`x = ( -3 \pm 2√(3))/( 3)`

`x = ( -3 + 2√(3))/( 3), \ ( -3 - 2√(3))/( 3)`