Quadratic Equation A3

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Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-a-3/ Fetched from algebrica.org test 3458; source modified 2025-03-06T16:14:07.

Solve the quadratic equation

-7x + 3 = -2x^2

First, we need to rewrite the second-degree equation in its standard form $ax^2+bx+c = 0$. By collecting all terms on the left side of the equal sign, we obtain::

2x^2-7x + 3 = 0

After reducing the equation to its standard form, we can substitute the coefficients $a=2, b=-7, c=3$ into the quadratic formula:

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

We obtain:

x_{1,2} = \frac{{-(-7) \pm \sqrt{{(-7)^2-4(2)(3)}}}}{{2(2)}}

In this case, the discriminant $\Delta$ is $\geq 0$ so the equation admits two distinct real solutions.

\begin{align*} x_{1,2} &= \frac{{7 \pm \sqrt{{49-24}}}}{4}\\[0.6em] &= \frac{{7 \pm \sqrt{{25}}}}{4}\\[0.6em] &= \frac{7 \pm 5}{4}\\[0.6em] \end{align*}

Finally, by performing the calculations, we obtain:

x_1=\frac{7 + 5}{4}=\frac{12}{4} = 3
x_2=\frac{7 - 5}{4}=\frac{2}{4} = \frac{1}{2}

The solution to the equation is:

x_1 = 3 \quad \quad x_2=\frac{1}{2}

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}