Ref.: module B

This exercise requires understanding the fundamental concepts of quadratic equation and their solution methods. It involves using the factorization method.

Solve the quadratic equation:

\[\sqrt{2}x^2 +7x +5\sqrt{2} = 0\]


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. \[\nexists \hspace{10px} x \in \mathbb{R}\]