Quadratic Equation A6

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Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-a-6/ Fetched from algebrica.org test 3470; source modified 2025-03-06T16:13:05.

Solve the quadratic equation:

(x - 4)^2 - 9 = 0

The term, $(x - 4)^2$, represents the square of a binomial of the form $(a-b)^2$ and can be expanded as $a^2-2ab+b^2$. Thus, the given equation can be rewritten as:

(x^2 - 8x+16)-9 = 0

After performing the necessary calculations, the equation can be written in the following form:

x^2 - 8x+7 = 0

We can substitute the coefficients $a= 1, b=-8, c=7$ into the quadratic formula:

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

We obtain:

x_{1,2} = \frac{{-(-8) \pm \sqrt{{(-8)^2 - 4(1)(7)}}}}{{2(1)}}

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

\begin{align*} x_{1,2} &= \frac{{8 \pm \sqrt{{64 - 28}}}}{2}\\[0.6em] &= \frac{{8 \pm \sqrt{{36}}}}{2} \end{align*}

Finally, by performing the calculations, we obtain:

x_1 = \frac{8 - 6}{2} = \frac{2}{2} = 1
x_2 = \frac{8 + 6}{2} = \frac{14}{2} = 7

The solution to the equation is:

x_1 =1 \quad x_2=7

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}