9x^2-3x-4=0 — step-by-step solution

1) Let's solve in

ℝ

9 x^{2} - 3 x - 4 = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

Δ = {(-3)}^{2} - 4 \cdot 9 \cdot (-4)

ℹ

{(-3)}^{2}

9

‖

4 \cdot 9

36

Δ = 9 + 36 \cdot 4

Δ = 9 + 144

Δ = 153

Δ > 0

, the equation has two solutions:

For the first solution, we obtain:

\frac{- b - \sqrt{Δ}}{2 a}

x1 = \frac{3 - \sqrt{153}}{2 \cdot 9}

ℹ

- \sqrt{153}

- 3 \cdot \sqrt{17}

‖

2 \cdot 9

18

x1 = \frac{3 - 3 \cdot \sqrt{17}}{18}

x1 = \frac{1 - \sqrt{17}}{6}

For the second solution, we obtain:

\frac{- b + \sqrt{Δ}}{2 a}

x2 = \frac{3 + \sqrt{153}}{2 \cdot 9}

ℹ

\sqrt{153}

3 \cdot \sqrt{17}

‖

2 \cdot 9

18

x2 = \frac{3 + 3 \cdot \sqrt{17}}{18}

x2 = \frac{1 + \sqrt{17}}{6}

Solution found:

x = \frac{1 - \sqrt{17}}{6}

Solution obtained:

x = \frac{\sqrt{17} + 1}{6}

▶Solution:

x = {\frac{1 - \sqrt{17}}{6}, \frac{\sqrt{17} + 1}{6}}

2) The numeric value of:

x = {\frac{1 - \sqrt{17}}{6}, \frac{\sqrt{17} + 1}{6}}

is:

x = {\frac{1 - \approx 4.1231}{6}, \frac{\approx 4.1231 + 1}{6}}

x = {- \frac{\approx 3.1231}{6}, \frac{\approx 5.1231}{6}}

▷

x = {\approx -0.5205, \approx 0.8539}