Question

If the expression \(\frac{1}{2}(x+c)(x-c)\), where \(c\) is a positive constant, can be rewritten as \(\frac{1}{2} x^2-5\), what is the value of \(c\) ? A) \(\sqrt{5}\) B) \(\sqrt{10}\) C) 5 D) 10

Short answer

The value of 'c' is \( \boxed{\text{(B)}\ \sqrt{10}} \).

Step-by-step solution

Expand the expression

First, expand the expression \(\frac{1}{2}(x+c)(x-c)\) using the distributive property (FOIL method): \(\frac{1}{2}(x^2 - c^2)\)

Compare the expanded expression with the given expression

We are given that the expanded expression is equal to \(\frac{1}{2} x^2-5\). Comparing the two expressions, we have: \(\frac{1}{2}(x^2 - c^2) = \frac{1}{2}x^2 - 5\)

Simplify and solve for 'c'

Now, we can see that the \(x^2\) terms on both sides will cancel out, leaving us with: \(-\frac{1}{2}c^2 = -5\) To solve for 'c', multiply both sides by \(-2\): \(c^2 = 10\) Now, take the square root of both sides. Since 'c' is a positive constant, we only consider the positive root: \(c = \sqrt{10}\) So the value of 'c' is \( \boxed{\text{(B)}\ \sqrt{10}} \).

Key concepts

Distributive Property

The distributive property is a fundamental algebraic principle that allows you to multiply a single term by each term inside a parenthesis separately. In simpler terms, it unfolds an expression that involves multiplication over addition or subtraction. For example, if you have a scenario like \(a(b + c)\), you would distribute \(a\) across both \(b\) and \(c\), resulting in \(ab + ac\).

When dealing with SAT math problems, the distributive property helps you to simplify complex expressions and solve equations efficiently. It's especially handy when you encounter polynomials and need to expand or factor them, as it's the case with the expression in our exercise \(\frac{1}{2}(x+c)(x-c)\).

FOIL Method

The FOIL method is a specific case of the distributive property used to multiply two binomials. FOIL stands for First, Outer, Inner, Last, representing the order in which you multiply the terms.

For instance, if we have \((a + b)(c + d)\), the FOIL method guides you to multiply the First terms \(a\) and \(c\), the Outer terms \(a\) and \(d\), the Inner terms \(b\) and \(c\), and the Last terms \(b\) and \(d\). This results in \(ac + ad + bc + bd\). In our exercise, the FOIL method is used to expand \((x+c)(x-c)\) into \(x^2 - c^2\), simplifying the initial expression.

Solving Quadratic Equations

Quadratic equations are a notch above linear equations and are recognizable by the \(x^2\) term, representing that \(x\) is squared. To solve a quadratic equation, you are basically finding the values of \(x\) that make the equation true.

There are multiple methods to solve quadratics - factoring, using the quadratic formula, completing the square, or graphically. In the SAT problem provided, we are already presented with a quadratic expression, set equal to another expression. We manipulate the equation algebraically to isolate \(c^2\) and solve for \(c\), ultimately determining the solution by taking the square root of both sides.

Square Root

The square root is an operation that asks: what number can be multiplied by itself to produce a given number? If we have \(x^2 = 9\), the square root of 9 is 3, because \(3^2\) or \(3 \times 3 = 9\).

The SAT problem solving step involves finding the square root \(c^2 = 10\). Since we're told that \(c\) is a positive constant, we only consider the positive square root, resulting in \(c = \sqrt{10}\). It's important to remember when taking the square root of both sides of an equation, there's typically a positive and a negative potential solution. However, context, like in this case specifying a positive constant, can dictate which root is appropriate.