Problem 52 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ x-5 \sqrt{x}+6=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 4\) and \(x = 9\).

Step 1: Identify the structure and start solving the equation

Let's observe the given equation \(x - 5\sqrt{x} + 6 = 0\). This equation has a combination of a linear and a square root term. To make this equation easier to solve, we can try using a substitution method. Let \(y = \sqrt{x}\), which implies \(x = y^2\). Substitute \(y\) into the equation to transform it.

Step 2: Apply substitution

After substituting \(y = \sqrt{x}\), the equation becomes \(y^2 - 5y + 6 = 0\). Now, we have a quadratic equation in terms of \(y\).

Step 3: Solve the quadratic equation

The quadratic equation \(y^2 - 5y + 6 = 0\) can be solved using factorization. We need two numbers whose product is 6 and sum is -5. The numbers -2 and -3 satisfy these conditions. Therefore, the equation factors as \((y-2)(y-3) = 0\).

Step 4: Find the values of y

Setting each factor to zero gives \(y - 2 = 0\) or \(y - 3 = 0\). Solving these, we get \(y = 2\) or \(y = 3\).

Step 5: Back-substitution to find x

Recall that \(y = \sqrt{x}\). Substituting back into the expression for \(x\), we have \(\sqrt{x} = 2\) and \(\sqrt{x} = 3\). Solving these gives \(x = 4\) and \(x = 9\).

Step 6: Verify the solutions

Ensure these solutions satisfy the original equation. Substitute \(x = 4\) into the original equation: \(4 - 5 \times 2 + 6 = 0\), which is true. Similarly, for \(x = 9\): \(9 - 5 \times 3 + 6 = 0\), which also holds true. Both solutions are verified.

Key Concepts

Quadratic Equation

A quadratic equation is any equation that can be rearranged in the standard form as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are numbers and \(a eq 0\). This is important because a zero in \(a\) would make it linear, not quadratic. Quadratic equations can have different types of solutions depending on the values of \(a\), \(b\), and \(c\). These could be real or complex solutions.Quadratic equations have certain properties:

• Discriminant: It helps determine the nature of the roots. Calculated as \(b^2 - 4ac\).
• Roots: The solutions for the variable. They can be found through various methods such as factoring, completing the square, and the quadratic formula.
• Graph: The graph of a quadratic equation is a parabola.

Most often, the roots of a quadratic equation need to be found, as these represent the solutions. In our exercise, after substitution, we ended with the simplified quadratic equation \(y^2 - 5y + 6 = 0\).

Substitution Method

The substitution method is a useful problem-solving tool that simplifies equations by replacing variables. It reduces complex expressions into a more manageable form. In our exercise, the given equation \(x - 5\sqrt{x} + 6 = 0\) contained a complex mix of linear and square root terms. To make it easier to solve, we performed a substitution:

• We let \(y = \sqrt{x}\). This simplified our original equation by converting it into a quadratic equation, \(y^2 - 5y + 6 = 0\).
• This transformation allows us to handle the square root term with basic algebra techniques for quadratics.

Once the new equations are solved for \(y\), we substitute back to find the original variable \(x\) by reversing the original substitution. It's a great way to tackle challenging problems by making them more approachable.

Square Root

The square root function is a fundamental concept in algebra. It allows us to find a number, which when squared, gives the original value. For example, the square root of 9 is 3 because \(3^2 = 9\).In our exercise, the square root serves as a bridge in the equation. The equation \(x - 5\sqrt{x} + 6 = 0\) includes \(\sqrt{x}\), which requires careful manipulation:

• To simplify, we used the substitution \(y = \sqrt{x}\), letting us transform the equation.
• Finding \(y\), we then take its square to determine the actual value of \(x\).

The ability to simplify and manipulate the square root is critical in solving such equations, especially when dealing with real solutions that must satisfy initial conditions. In this problem, after finding \(y\), we solved \(\sqrt{x} = 2\) and \(\sqrt{x} = 3\) to arrive at the real solutions for \(x\): \(x = 4\) and \(x = 9\). This exemplifies how square roots and their manipulation are integral to solving equations with variable square roots.