Problem 57 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the equation. $$ \sqrt{u^{2}+u-5}=1 $$

Step-by-Step Solution

Answer

The equation \(\sqrt{u^2 + u - 5} = 1\) can be solved by following these steps: 1. Square both sides: \(u^2 + u - 5 = 1\) 2. Bring all terms to one side: \(u^2 + u - 6 = 0\) 3. Factor the quadratic expression: \((u + 3)(u - 2) = 0\) 4. Solve for u: \(u = -3, 2\)

Step 1: Square both sides of the equation

In order to eliminate the square root, we will square both sides of the equation: \[\left(\sqrt{u^2 + u - 5}\right)^2 = 1^2\]

Step 2: Simplify the equation

Squaring both sides to simplify the expression: \[u^2 + u - 5 = 1\]

Step 3: Bringing all terms to one side

To proceed, we have to move all the terms to one side of the equation: \[u^2 + u - 6 = 0\]

Step 4: Factoring the quadratic expression

Now we will factor the quadratic expression: \[(u+3)(u-2) = 0\]

Step 5: Solve for u

Now we will solve for u by setting each factor equal to zero: \[u+3 = 0 \Rightarrow u = -3\] \[u-2 = 0 \Rightarrow u = 2\] So, the solutions for the equation are \(u = -3\) and \(u = 2\).

Key Concepts

Factoring Quadratics

Factoring quadratics is a fundamental technique in solving quadratic equations, which are expressed in the form of ax^2 + bx + c = 0. In this method, we seek to break down the quadratic equation into a product of two binomials. The key idea behind this approach is to find two numbers that both add up to the coefficient of the middle term (b) and multiply together to equal the product of the first coefficient (a) and the constant term (c).

Once we have those two numbers, we rewrite the equation in the factored form \((x+n)(x+m)=0\) such that:

• The sum of n and m equals to b.
• The product of n and m equals to a * c.

Applying the Zero Product Property, we set each binomial equal to zero and solve for x, which gives us the possible solutions to the quadratic equation. In the given exercise, after rearranging the terms, the equation transformed into \(u^2 + u - 6 = 0\), which upon factoring becomes \((u+3)(u-2)=0\).Due to the Zero Product Property, we know one of the factors must equal zero, which leads us to the solutions \(u = -3\) and \(u = 2\).

Square Root Equations

When dealing with square root equations, where the variable is under a square root, the primary step is to isolate the square root expression on one side of the equation. The next phase is to eliminate the square root by raising both sides of the equation to the power of two, effectively squaring the equation. This step must be handled with care, as squaring both sides of an equation can sometimes introduce extraneous solutions that don't actually satisfy the original equation.

After the square root has been eliminated, we are often left with a quadratic equation, which will then require us to use another method, such as factoring, completing the square, or using the quadratic formula, to find the solutions. In our initial problem \(\sqrt{u^2 + u - 5} = 1\), squaring both sides provided us with a quadratic equation. It is crucial at this point to check any potential solutions against the original equation, as the squaring process might include solutions that make no sense when plugged back in.

Quadratic Formula

The quadratic formula is a powerful tool for solving any quadratic equation, even when factoring is cumbersome or impossible to perform. The formula is derived from the standard quadratic equation ax^2+bx+c=0 and is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• The term \(b^2 - 4ac\) is called the discriminant and it tells us about the nature of the roots. A positive discriminant indicates two real and distinct solutions, zero means that there is a single, repeated real solution, and a negative discriminant implies complex (non-real) solutions.
• The symbols \(\pm\) suggest that there are two solutions, one from using a positive sign and the other from using a negative sign.

Using the quadratic formula is straightforward: replace a, b, and c with the values from your equation and calculate. It is especially useful when factoring does not easily reveal the solutions or when the equation is not factorable by integers. In cases where we can factor the equation, as with our example \(u^2 + u - 6 = 0\), using the quadratic formula would provide the same solutions obtained by factoring: \(u = -3\) and \(u = 2\).