Quadratic Equation B9

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Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-b-9/ Fetched from algebrica.org test 5010; source modified 2025-03-06T18:07:34.

Solve the quadratic equations using the factorization method.

\sqrt{2}x^2 +7x +5\sqrt{2} = 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 = (7)^2 - 4(\sqrt{2})(5) = 49 + 20\sqrt{2} \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 $b = 7$ and whose product $P = r_1 \cdot r_2$ equals $a \cdot c = \sqrt{2} \cdot 5\sqrt{2}= 10$. 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 & 5 & 10 & 7 \\\\ & -2 & -5 & 10 & -7\\\\ \end{array}

The numbers $r_1, r_2$ satisfying the constraint are 2 and 5. Then we need to rewrite the polynomial as $ax^2 + r_{1}x + r_{2}x + c$.

The equation becomes:

\sqrt{2}x^2+2x++5x+5\sqrt{2} = 0

Factoring common terms, we get:

\begin{align*} & \sqrt{2}x^2+2x+5(x+\sqrt{2}) = 0 \\[1em] & \sqrt{2}x^2+\sqrt{2}\cdot\sqrt{2}x^2x+5(x+\sqrt{2}) = 0 \\[1em] & \sqrt{2}x(x+\sqrt{2})+5(x+\sqrt{2}) = 0 \\[1em] & (\sqrt{2}x+5)(x+\sqrt{2}) = 0 \end{align*}

The solutions are the values of $x$ for which $\sqrt{2}x+5 = 0$ and $x+\sqrt{2} = 0$.

\sqrt{2}x+5 = 0 \to \sqrt{2}x+5 \to \sqrt{2}x = -5 \to x = -\frac{5}{\sqrt{2}}
x+\sqrt{2} = 0 \to x=-\sqrt{2}

The solution to the equation is:

The solution to the equation is:

x=-\frac{5}{\sqrt{2}} \quad\quad x=-\sqrt{2}

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}