Chapter 25: Problem 50
What is the value of \(x^2-y^2\) ? (1) \(x+y=12\) (2) \(x y=35\)
Short Answer
Expert verified
The value of \(x^2 - y^2\) is 24.
Step by step solution
01
Understand the Problem
We need to find the value of the expression \(x^2 - y^2\). The problem provides us with two equations: \(x + y = 12\) and \(xy = 35\).
02
Recognize the Algebraic Identity
Notice that \(x^2 - y^2\) can be expressed as a difference of squares: \(x^2 - y^2 = (x+y)(x-y)\). This will simplify our solution process.
03
Assign Known Values
From statement (1), we know that \(x + y = 12\). Thus, part of the identity \((x+y)(x-y)\) is given as \(12\).
04
Solve for \(x-y\)
To find \(x-y\), we need both \(x+y\) and \(xy\). We'll use the fact that \(x\) and \(y\) are roots of the quadratic equation \(t^2 - (x+y)t + xy = 0\). This becomes \(t^2 - 12t + 35 = 0\).
05
Calculate Roots
Find the roots of the quadratic \(t^2 - 12t + 35 = 0\) using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -12\), and \(c = 35\).
06
Evaluate the Quadratic Formula
Calculate \(t = \frac{12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 35}}{2}\) which simplifies to \(t = \frac{12 \pm \sqrt{144 - 140}}{2}\), or \(t = \frac{12 \pm \sqrt{4}}{2}\).
07
Determine the Values of \(x\) and \(y\)
The roots are \(t = \frac{12 + 2}{2} = 7\) and \(t = \frac{12 - 2}{2} = 5\). Thus, \(x = 7\) and \(y = 5\) or vice versa.
08
Compute \(x-y\)
Since \(x = 7\) and \(y = 5\), then \(x-y = 7 - 5 = 2\).
09
Substitute into the Identity
Plug the values into the identity: \(x^2 - y^2 = (x+y)(x-y) = 12 \cdot 2 = 24\).
10
State the Final Answer
The value of \(x^2 - y^2\) is \(24\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations are essential parts of algebra and appear frequently in mathematics. A standard quadratic equation has the form \(at^2 + bt + c = 0\), where \(a\), \(b\), and \(c\) are constants. The solutions to these equations, called roots, can be found using various methods, one of which is the quadratic formula:
- \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Difference of Squares
The difference of squares is a powerful algebraic identity that simplifies the process of solving equations like \(x^2 - y^2\). It can be expressed as:
It allows us to substitute directly into
- \(x^2 - y^2 = (x+y)(x-y)\).
It allows us to substitute directly into
- \((x+y)(x-y)\)
Expression Simplification
Simplifying expressions is a crucial skill in algebra, allowing one to reduce complex expressions into more manageable forms. In the given exercise, once we've recognized the identity \(x^2 - y^2 = (x+y)(x-y)\), we substitute known values into this simple expression.
From the steps, we know:
From the steps, we know:
- \(x+y=12\)
- \(x-y=2\)
- Calculate: \((x+y)(x-y) = 12 \times 2 = 24\).