Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-b-3/ Fetched from algebrica.org test 4583; source modified 2025-03-06T18:04:27.
Solve the quadratic equations using the factorization method.
The equations is in the standard form $ax^2+bx+c=0$. First, it is essential to verify the its discriminant $\Delta = b^2 - 4ac$ is $\geq0 \quad$ to ensure the equation admits solutions in the field of real numbers. Substituting the coefficients of the equation into $\Delta$, we get:
$\Delta \gt 0$ means the equation has real solutions.
Now, we need to factorize the [polynomial]…/…/polynomials/). We must find two numbers, $r_1, r_2$ whose sum $S = r_1 + r_2$ equals $b = -4$ and whose product $P= r_1 \cdot r_2$ equals $a \cdot c = 3 \cdot -15 = -45$. We can use this simple table to find the numbers that satisfy our constraints.
- The numbers $r_1, r_2$ satisfying the constraint are 5 and -9 (row 1).We can rewrite the polynomial as $ax^2 + r_{1}x + r_{2}x + c$.
The equation becomes:
Factoring common terms, we get:
The solutions are the values of $x$ for which $3x+5 = 0$ and $x-3 = 0$.
The solution to the equation is:
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.
- If $b^2 - 4ac = 0$, the quadratic equation has two coincident real solutions.
- If $b^2 - 4ac = < 0$, the quadratic equation has no real solutions. Instead, it gives rise to complex solutions characterized by imaginary components.