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Agata [3.3K]
2 years ago
6

Pllllllllllllllllllllleasee one guys i neeed ur help one

Mathematics
2 answers:
Romashka [77]2 years ago
7 0

Answer:

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

Explanation:

<u>Given expression</u>:

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}

Svetllana [295]2 years ago
5 0

{ \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}

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