Problem 46 - Algebra and Trigonometry

Question

Find all real solutions of the equation. \(25 x^{2}+70 x+49=0\)

Step-by-Step Solution

Answer

The real solution is \( x = -\frac{7}{5} \).

Step 1: Identify the Quadratic Equation

Recognize that the equation is a quadratic equation in the standard form: \( ax^2 + bx + c = 0 \). Here, \( a = 25 \), \( b = 70 \), and \( c = 49 \).

Step 2: Check if the Quadratic is a Perfect Square

Expand \( (5x + 7)^2 \) to see if it matches the original equation. Calculate \((5x + 7)^2 = 25x^2 + 70x + 49\). Upon expansion, it is exactly the original equation.

Step 3: Solve the Perfect Square Equation

Since \( (5x + 7)^2 = 0 \), solve for \( x \) by setting the expression to zero: \( 5x + 7 = 0 \).

Step 4: Isolate x

Subtract 7 from both sides to get: \( 5x = -7 \).

Step 5: Solve for x

Divide each side by 5 to solve for \( x \): \( x = -\frac{7}{5} \).

Key Concepts

Perfect Square in Quadratic Equations

Quadratic equations often become simpler when they are a perfect square. A perfect square in this context means that the equation can be expressed as a square of a binomial. This occurs when the quadratic can be rewritten in the form \( (px + q)^2 = 0 \), indicating it has a double root. This is a useful form because it directly reveals the solution. To recognize a perfect square, factor the quadratic as if it were \( (px + q)^2 \) and expand it to see if it matches the original equation.

For instance, the equation \( 25x^2 + 70x + 49 = 0 \) matches \( (5x + 7)^2 = 0 \) upon expansion, confirming it is a perfect square. This simplification helps to see the structure and ultimately solve the equation quickly by setting the inside of the square to zero: \( 5x + 7 = 0 \).

• Identify if the quadratic can be expressed in the form \( (px + q)^2 \).
• Expand \( (px + q)^2 \) to verify it matches the original equation.
• Solve \( px + q = 0 \) to find \( x \).

Real Solutions of Quadratic Equations

Real solutions of quadratic equations are the values that satisfy the equation and can be plotted on a real-number line. When a quadratic is a perfect square, it only has one real solution, often called a double root. This means the parabola of the equation just touches the x-axis without crossing it.

In our example, after identifying that \( 25x^2 + 70x + 49 = 0 \) is a perfect square, we found a single real solution: \( x = -\frac{7}{5} \). This solution represents the point where the parabola reaches its vertex, touching the x-axis without crossing:

• If an equation is a perfect square, it has one real solution.
• The solution \( x \) can be found by setting the binomial to zero and solving.
• The graph of the equation only touches the x-axis at its vertex point.

Steps for Solving Quadratic Equations

Solving quadratic equations typically involves several steps, whether or not they are perfect squares. Here's a simple guide for typical cases:

First, ensure your equation is in the standard quadratic form \( ax^2 + bx + c = 0 \). Next, determine if the quadratic can be simplified by factoring, using the quadratic formula, or discovering it is a perfect square. In our specific case where we dealt with \( 25x^2 + 70x + 49 = 0 \), recognizing the quadratic as a perfect square sped up the solving process.

Here are the usual steps simplified:

• Identify standard quadratic form \( ax^2 + bx + c = 0 \).
• Check if the equation is a perfect square or can be factored easily.
• If not directly factorable or a perfect square, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \).
• Perform calculations to find real solutions, taking care to interpret any complex or repeated results appropriately.

By understanding these core practices, you can handle any quadratic equation confidently and effectively.