Problem 35 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ (x+5)^{2}-3(x+5)-10=0 $$

Step-by-Step Solution

Answer

The real solutions are \( x = 0 \) and \( x = -7 \).

Step 1: Substitute a new variable

To simplify the equation, substitute a new variable \( y = x + 5 \). This gives us the equation \( y^2 - 3y - 10 = 0 \).

Step 2: Identify the quadratic form

The equation \( y^2 - 3y - 10 = 0 \) is a quadratic equation in the standard form \( ay^2 + by + c = 0 \) with \( a = 1 \), \( b = -3 \), and \( c = -10 \).

Step 3: Apply the quadratic formula

Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( y \). Substituting \( a = 1 \), \( b = -3 \), and \( c = -10 \), we have: \[y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 40}}{2} \]

Step 4: Calculate the discriminant

Calculate the discriminant: \( 9 + 40 = 49\). Since the discriminant is positive, there are two real solutions.

Step 5: Solve for 'y' using the quadratic formula

Substitute back into the quadratic formula to find the solutions for \( y \): \[y = \frac{3 \pm 7}{2}\]This yields two solutions: \( y = \frac{3 + 7}{2} = 5 \) and \( y = \frac{3 - 7}{2} = -2 \).

Step 6: Substitute back to find 'x'

Recall that \( y = x + 5 \), so we'll solve for \( x \) for both values of \( y \):1. If \( y = 5 \), then \( x + 5 = 5 \), giving \( x = 0 \).2. If \( y = -2 \), then \( x + 5 = -2 \), giving \( x = -7 \).

Step 7: Conclude the real solutions

The real solutions for the equation \( (x+5)^{2}-3(x+5)-10=0 \) are \( x = 0 \) and \( x = -7 \).

Key Concepts

Substitution Method

The substitution method in quadratic equations involves introducing a new variable to simplify the problem. Consider a situation where you're dealing with a complex expression, like

• \((x+5)^{2}-3(x+5)-10=0\) .

To make the problem easier, you can substitute the expression inside the parentheses with a new variable. So, let's set \( y = x+5 \). This transforms our equation into

• \( y^2 - 3y - 10 = 0 \).

This substitution is very useful because it turns what might have been cumbersome algebra into a straightforward quadratic equation. Once you solve for \( y \) using the quadratic methods, you can simply substitute back the value of the original variable. It’s like unwrapping a complicated gift with ease to find the answers inside.

Quadratic Formula

The quadratic formula is a powerful tool used to find solutions to quadratic equations. These equations are in the form \( ax^2 + bx + c = 0 \) and their solutions can be calculated with the formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

This formula is effective because it provides solutions for any quadratic equation, even when the roots are not easily factorable. For example, substituting \( a = 1 \), \( b = -3 \), and \( c = -10 \) into the formula, we find

• \( y = \frac{3 \pm \sqrt{49}}{2} \).

This yields the roots for \( y \), giving us a direct path to the solutions. What’s great about the quadratic formula is its reliability – once you have the numbers to substitute, the formula does the hard work for you.

Real Solutions

Real solutions in the context of quadratic equations refer to solutions that are actual numbers on the real number line, as opposed to imaginary numbers. When you apply the quadratic formula, the nature of the solutions depends greatly on what's under the square root – that is, the discriminant. If the discriminant is positive, there are two different real solutions. If it is zero, there is exactly one real solution (a repeated root).

• If the discriminant is negative, the solutions are not real; they are complex or imaginary.

In our equation, the discriminant was positive, leading to two real solutions for \( y \):

• \( y = 5 \)
• \( y = -2 \)

These solutions confirm that our original substitution successfully simplified the problem.

Discriminant

The discriminant is a vital part of the quadratic formula, found beneath the square root sign. It is the expression \( b^2 - 4ac \). This little component tells you a lot about the nature of the equation's solutions:

• If the discriminant is positive, expect two distinct real solutions.
• If it is zero, only one real solution exists, indicating that the entity has a double root.
• If it's negative, then the solutions are not on the real number line but instead involve complex numbers.

In the exercise given, the discriminant calculated to be \( 49 \), which is positive. This signified that there would be two real solutions. Understanding the discriminant is like holding a key to unlock the secrets of your quadratic equation's solutions.

Standard Form of Quadratic Equations

Quadratic equations are typically written in their standard form: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. This form is crucial because it allows us to apply the quadratic formula directly.

• Once in this form, coefficients \( a \), \( b \), and \( c \) can be easily identified, allowing functions such as the quadratic formula to work.

These coefficients make it easy to apply further methods, like factoring or completing the square, should they simplify the solving process.

In our problem, after substituting \( y = x+5 \), the equation \( y^2 - 3y - 10 = 0 \) is already in standard form where \( a = 1 \), \( b = -3 \), and \( c = -10 \). Without getting the equation into this form, using tools like the quadratic formula becomes complicated.