Problem 26 - Algebra and Trigonometry

Question

\(19-32\) . Find all solutions of the system of equations. $$ \left\\{\begin{array}{l}{x+\sqrt{y}=0} \\ {y^{2}-4 x^{2}=12}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \((x, y) = (-\sqrt{6}, 6)\).

Step 1: Express x in terms of y

From the equation \(x + \sqrt{y} = 0\), solve for \(x\):\[ x = -\sqrt{y} \]

Step 2: Substitute x in the second equation

Substitute \(x = -\sqrt{y}\) into the second equation \(y^{2} - 4x^{2} = 12\):\[ y^{2} - 4(-\sqrt{y})^{2} = 12 \] Simplify this to get: \[ y^{2} - 4y = 12 \]

Step 3: Solve for y

Rearrange the equation \(y^{2} - 4y = 12\) to form a quadratic equation:\[ y^{2} - 4y - 12 = 0 \] Solve the quadratic equation using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), \(c = -12\):\[ y = \frac{4 \pm \sqrt{16 + 48}}{2} \]\[ y = \frac{4 \pm 8}{2} \] This results in two solutions: \(y = 6\) or \(y = -2\).

Step 4: Discard invalid y values

Since \(y = -2\) leads to taking a square root of a negative number which is not real, we discard it. Thus, only \(y = 6\) is possible.

Step 5: Find x using the valid y value

Using \(y = 6\), substitute back into \(x = -\sqrt{y}\) to find \(x\):\[ x = -\sqrt{6} \] Hence, the solution is \((x, y) = (-\sqrt{6}, 6)\).

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of degree two. It typically takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratic equations can be solved using various methods, including factoring, completing the square, and the quadratic formula.

The quadratic formula is a widely used tool because it provides a general solution for all quadratic equations. The formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula calculates the values of \( x \) by finding the points at which the quadratic equation intersects the \( x \)-axis.

In the exercise, rearranging the equation \( y^2 - 4y - 12 = 0 \) into standard quadratic form allows us to easily identify \( a = 1 \), \( b = -4 \), and \( c = -12 \). Using these values in the quadratic formula, we are able to solve for \( y \) and get possible solutions: \( y = 6 \) and \( y = -2 \).

Real Solutions

Real solutions of an equation are the solutions that can be represented on the real number line. In the context of quadratic equations, real solutions will have real and numeric values.

The discriminant \( b^2 - 4ac \) determines the nature of the roots for a quadratic equation:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution (a repeated root).
• If the discriminant is negative, there are no real solutions, but two complex solutions.

In our exercise, the discriminant \( 16 + 48 = 64 \) is positive, implying two potential real solutions, \( y = 6 \) and \( y = -2 \). However, only \( y = 6 \) is valid as it leads to a real number when substituted for \( x \), because the square root operation requires a non-negative input. Hence, the real solution set is \((x, y) = (-\sqrt{6}, 6)\).

Substitution Method

The substitution method is a technique used to solve systems of equations by substituting one equation into the other. This method is particularly useful for solving systems where one equation can be easily rearranged to express one variable in terms of the other.

In the given exercise, we start with the equation \( x + \sqrt{y} = 0 \). We solve for \( x \) by isolating it: \( x = -\sqrt{y} \).

Next, we substitute this expression for \( x \) in the second equation \( y^2 - 4x^2 = 12 \). This substitution reduces the system of equations into a single quadratic equation in terms of \( y \), specifically \( y^2 - 4y = 12 \), which can then be solved using methods for solving quadratic equations.

Once the valid solution for \( y \) is found, it is substituted back to find \( x \), ensuring consistency and correctness in the solution. This method efficiently narrows down the number of potential solutions, simplifying the process of solving the system.