Quadratic Equation B10

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Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-b-10/ Fetched from algebrica.org test 5012; source modified 2025-03-06T18:08:06.

Solve the quadratic equations using the factorization method.

12x^2 +17x - 5 = 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 = (17)^2 - 4(12)(-5) = 529 \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 $b = 17$ and whose product $P = r_1 \cdot r_2$ equals $a \cdot c = 12 \cdot -5= -60$. We can use this simple table to find the numbers that satisfy our constraints. $-60$ can be factored into 20 and - 3 and that the sum of the two numbers gives 17.


The equations becomes:

12x^2 +20x -3x - 5 = 0

Factoring the common terms, we get: \begin{align*} & 12x^2 -3x +20x- 5 = 0 \[0.6em] & 3x(4x-1) + 5(4x-1) = 0 \[0.6em] & (3x + 5)(4x-1) = 0 \end{align*}

The solutions are the values of $x$ for which $3x + 5 = 0$ and $4x-1 = 0$

3x + 5 = 0 \to 3x=-5 \to x = -\frac{5}{3}
4x-1 = 0 \to 4x=1 \to x = \frac{1}{4}

The solution to the equation is:

x = -\frac{5}{3} \quad \quad x = \frac{1}{4}

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}