Quadratic Equation B7

Copy Markdown View Source

Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-b-7/ Fetched from algebrica.org test 4600; source modified 2025-03-06T18:06:28.

Solve the quadratic equations using the factorization method.

(4x + 8)\left(\frac{1}{2}x-6\right) = 0

Expanding the equation using the distributive property, we get:

\begin{align*} & \frac{4}{2}x^2-24x + \frac{8}{2}x-48 = 0 \\[0.6em] &2x^2-24x + 4x - 48 = 0\\[1em] &2x^2-20x-48 = 0 \end{align*}

The coefficients $a, b$ and $c$ have 2 as common multiplier. We can simplify the equation which is now brought into the standard form of a quadratic equation:

x^2-10x-24 = 0

The equations is in the standard form $ax^2+bx+c=0$. It is essential to verify the its discriminant $\Delta = b^2 - 4ac$ is $\geq0$ to ensure the equation admits solutions in the field of real numbers. Substituting the coefficients of the equation into $\Delta$, we get:

\Delta = (-10)^2 - 4(1)(-26) = 100 + 96 = 196 \geq 0

$\Delta \gt 0$ means the equation has real solutions.

Now, we need to factorize the polynomial. We must find two numbers, $r_1, r_2$ whose sum $S = r_1 + r_2$ equals $b = -10$ and whose product $P = r_1 \cdot r_2$ equals $a \cdot c = 1 \cdot -24 = -24$. We can use this simple table to find the numbers that satisfy our constraints.

\begin{array}{rrrr} & r_1 & r_2 & P & S \\\\ \hline & 2 & - 12 & -24 & -10 \\\\ & -2 & +12 & -24 & 10\\\\ \end{array}

The numbers $r_1, r_2$ satisfying the constraint are $2$ and $-12$ (row 1). Then we need to rewrite the polynomial as $ax^2 + r_{1}x + r_{2}x + c$.

The equation becomes:

x^2+2x-12x-24 = 0

Factoring common terms, we get:

\begin{align*} &x(x+2)-12(x+2) = 0 \\[0.6em] &(x+2)(x-12) = 0 \end{align*}

The solutions are the values of $x$ for which $x-2= 0$ and $x-12 = 0$.

x -2 = 0 \to x=-2
x -12 = 0 \to x=12

The solution to the equation is:

The solution to the equation is:

x_=-2 \quad \quad x_1 =12

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}