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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. We must find two numbers, $r_1, r_2$ whose sum $S = r_1 + r_2$ equals $b = -7$ and whose product $P= r_1 \cdot r_2$ equals $a \cdot c = 6$. We can use this simple table to find the numbers that satisfy our constraints.
The numbers $r_1, r_2$ satisfying the constraint are $-1$ and $-6$ (row 2). 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 $2x-1 = 0$ 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.