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Let's solve ~

Calculate discriminant :

[tex]\qquad \sf  \dashrightarrow \: 3 {x}^{2} + 6x - 1[/tex]

  • a = 3
  • b = 6
  • c = 1

[tex]\qquad \sf  \dashrightarrow \: discriminant = {b}^{2} - 4ac[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = (6) {}^{2} - (4 \times 3 \times 1)[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = 36 - 12[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = 24[/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt {d} = 2 \sqrt{6} [/tex]

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

[tex]\qquad \sf  \dashrightarrow \: x = \cfrac{ - b \pm \sqrt{d} }{2a} [/tex]

[tex]\qquad \sf  \dashrightarrow \: x = \cfrac{ - 6\pm 2 \sqrt{6} }{2 \times 3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: x = \cfrac{ - 6\pm 2 \sqrt{6} }{6} [/tex]

So, the intercepts are :

[tex]\qquad \sf  \dashrightarrow \: x = \cfrac{ - 6 - 2 \sqrt{6} }{6} [/tex]

and

[tex]\qquad \sf  \dashrightarrow \: x = \cfrac{ - 6 + 2 \sqrt{6} }{6} [/tex]

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Answer:

[tex]\left( \dfrac{ -3 + 2\sqrt{3}}{ 3}, \ 0\right), \ \left(\dfrac{ -3 - 2\sqrt{3}}{ 3}, \ 0\right)[/tex]

Explanation:

Given expression:

f(x) = 3x² + 6x - 1

  • To find x intercepts, set f(x) = 0

Use quadratic formula:

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

Here after finding coefficients:

  • a = 3, b = 6, c = -1

Applying formula:

[tex]x = \dfrac{ -6 \pm \sqrt{6^2 - 4(3)(-1)}}{2(3)}[/tex]

[tex]x = \dfrac{ -6 \pm \sqrt{48}}{6}[/tex]

[tex]x = \dfrac{ -6 \pm 4\sqrt{3}}{6}[/tex]

[tex]x = \dfrac{ -6 \pm 4\sqrt{3}}{2 \cdot 3}[/tex]

[tex]x = \dfrac{ -3 \pm 2\sqrt{3}}{ 3}[/tex]

[tex]x = \dfrac{ -3 + 2\sqrt{3}}{ 3}, \ \dfrac{ -3 - 2\sqrt{3}}{ 3}[/tex]

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