Question

Pllllllllllllllllllllleasee one guys i neeed ur help one

Answer 1

Answer:

$\left( \dfrac{ -3 + 2\sqrt{3}}{ 3}, \ 0\right), \ \left(\dfrac{ -3 - 2\sqrt{3}}{ 3}, \ 0\right)$

Explanation:

Given expression:

f(x) = 3x² + 6x - 1

• To find x intercepts, set f(x) = 0

Use quadratic formula:

$\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \ where \ ax^2 + bx + c = 0$

Here after finding coefficients:

• a = 3, b = 6, c = -1

Applying formula:

$x = \dfrac{ -6 \pm \sqrt{6^2 - 4(3)(-1)}}{2(3)}$

$x = \dfrac{ -6 \pm \sqrt{48}}{6}$

$x = \dfrac{ -6 \pm 4\sqrt{3}}{6}$

$x = \dfrac{ -6 \pm 4\sqrt{3}}{2 \cdot 3}$

$x = \dfrac{ -3 \pm 2\sqrt{3}}{ 3}$

$x = \dfrac{ -3 + 2\sqrt{3}}{ 3}, \ \dfrac{ -3 - 2\sqrt{3}}{ 3}$

Answer 2

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

Let's solve ~

Calculate discriminant :

$\qquad \sf \dashrightarrow \: 3 {x}^{2} + 6x - 1$

• a = 3
• b = 6
• c = 1

$\qquad \sf \dashrightarrow \: discriminant = {b}^{2} - 4ac$

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

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

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

$\qquad \sf \dashrightarrow \: \sqrt {d} = 2 \sqrt{6}$

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

$\qquad \sf \dashrightarrow \: x = \cfrac{ - b \pm \sqrt{d} }{2a}$

$\qquad \sf \dashrightarrow \: x = \cfrac{ - 6\pm 2 \sqrt{6} }{2 \times 3}$

$\qquad \sf \dashrightarrow \: x = \cfrac{ - 6\pm 2 \sqrt{6} }{6}$

So, the intercepts are :

$\qquad \sf \dashrightarrow \: x = \cfrac{ - 6 - 2 \sqrt{6} }{6}$

and

$\qquad \sf \dashrightarrow \: x = \cfrac{ - 6 + 2 \sqrt{6} }{6}$