Quadratic Equation A8

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Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-a-8/ Fetched from algebrica.org test 3482; source modified 2025-03-06T16:12:23.

Solve the quadratic equation:

x^2 + 0.4x - 0.16 = 0

To solve the equation, we can get rid of the decimals by multiplying every coefficient by 100, obtaining:

100x^2 + 40x - 16 = 0

The equation is already reduced to the standard form $ax^2+bx+c= 0$. We can substitute the coefficients $a=2, b=10, c=11$ into the quadratic formula:

x_{1,2} = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}

We obtain:

\begin{align*} x_{1,2} &= \frac{-40 \pm \sqrt{40^2-4(100)(-16)}}{2(100)}\\[0.8em] &= \frac{-40 \pm \sqrt{1600+6400}}{200} \end{align*}

In this case, the discriminant $\Delta$ is $\geq 0$ so the equation admits two distinct real solutions. by performing the calculations we have:

\begin{align*} x_{1,2} &= \frac{{-40\pm \sqrt{{8000}}}}{200}\\[0.8em] &= \frac{{-40\pm \sqrt{{1600 \times 5}}}}{200}\\[0.8em] &= \frac{{-40\pm \sqrt{{40^2 \times 5}}}}{200}\\[0.8em] &= \frac{{-40\pm 40\sqrt{{5}}}}{200}\\[0.8em] &= \frac{{-1 \pm \sqrt{{5}}}}{5} \end{align*}

Finally, we obtain:

x_1 = \frac{{-1 + \sqrt{{5}}}}{5}
x_2= \frac{{-1-\sqrt{{5}}}}{5}

The solution to the equation is:

x = \frac{{-1+\sqrt{{5}}}}{5} \quad x = \frac{{-1-\sqrt{{5}}}}{5}

Remember that the discriminant is crucial in determining the nature and number of solutions of quadratic equations.

  • If $b^2 - 4ac > 0$, the quadratic equation has two distinct real solutions.
S = \{x_1, x_2\} \quad x_1, x_2 \in \mathbb{R} \quad x_1 \neq x_2
  • If $b^2 - 4ac = 0$, the quadratic equation has two coincident real solutions.
S = \{x\} \quad x \in \mathbb{R} \quad x = x_1 = x_2
  • If $b^2 - 4ac = < 0$, the quadratic equation has no real solutions. Instead, it gives rise to complex solutions characterized by imaginary components.
\nexists \hspace{10px} x \in \mathbb{R}