Chapter 10: Problem 34
Solve. $$\begin{aligned} &p+q=-4\\\ &p q=-5 \end{aligned}$$
Short Answer
Expert verified
The solutions are \(p = -5\), \(q = 1\) or \(p = 1\), \(q = -5\).
Step by step solution
01
Understand the problem
Given a system of equations: \(\begin{aligned} &p + q = -4 \ &pq = -5 \end{aligned}\) These equations represent the sum and product of two numbers, respectively.
02
Write the quadratic equation
From the equations, form a quadratic equation whose roots are the values of \(p\) and \(q\): \(x^2 - (p+q)x + pq = 0\)
03
Substitute given values
Substitute \(p + q = -4\) and \(pq = -5\) into the quadratic equation: \(x^2 - (-4)x + (-5) = 0\) This simplifies to \(x^2 + 4x - 5 = 0\)
04
Factorize the quadratic equation
Factorize the quadratic equation \(x^2 + 4x - 5 = 0\) into two binomials: \((x + 5)(x - 1) = 0\)
05
Solve for \(x\)
Set each factor equal to zero and solve for \(x\): \(x + 5 = 0\) or \(x - 1 = 0\) Thus, \(x = -5\) and \(x = 1\)
06
Identify the solutions
The solutions to the original system are \(p = -5\) and \(q = 1\), or \(p = 1\) and \(q = -5\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
System of Equations
A system of equations consists of two or more equations with the same set of variables. Solving a system means finding all possible values for the variables that satisfy all given equations. In our example, we have the system:
\(p + q = -4\) and \(pq = -5\).
This system tells us two things: the sum of two numbers \(p\) and \(q\) is \(-4\), and their product is \(-5\).
Solving this system will help us find the specific values of \(p\) and \(q\) that meet both conditions.
\(p + q = -4\) and \(pq = -5\).
This system tells us two things: the sum of two numbers \(p\) and \(q\) is \(-4\), and their product is \(-5\).
Solving this system will help us find the specific values of \(p\) and \(q\) that meet both conditions.
Factoring Quadratic Equations
Factoring quadratic equations is a method used to solve quadratics by breaking them into simpler expressions, or 'factors'. The standard quadratic equation is in the form:
\(ax^2 + bx + c = 0\).
In our problem, we derived the quadratic equation
\(x^2 + 4x - 5 = 0\).
To factorize, look for two numbers that multiply to give \(c\) (in this case, \(-5\)) and add to give \(b\) (which is \(4\)).
In this case, \(5\) and \(-1\) work because they satisfy both conditions. Therefore, the factored form is:
\((x + 5)(x - 1) = 0\).
\(ax^2 + bx + c = 0\).
In our problem, we derived the quadratic equation
\(x^2 + 4x - 5 = 0\).
To factorize, look for two numbers that multiply to give \(c\) (in this case, \(-5\)) and add to give \(b\) (which is \(4\)).
In this case, \(5\) and \(-1\) work because they satisfy both conditions. Therefore, the factored form is:
\((x + 5)(x - 1) = 0\).
Roots of Equations
The roots of an equation are the values of the variables that make the equation true. In the factored form,
\((x + 5)(x - 1) = 0\),
we set each factor equal to 0 to find the roots:
\(x + 5 = 0\) and \(x - 1 = 0\).
Solving these gives:
\(x = -5\) and \(x = 1\).
These values are the solutions or roots of the quadratic equation. They represent the possible values for \(p\) and \(q\).
\((x + 5)(x - 1) = 0\),
we set each factor equal to 0 to find the roots:
\(x + 5 = 0\) and \(x - 1 = 0\).
Solving these gives:
\(x = -5\) and \(x = 1\).
These values are the solutions or roots of the quadratic equation. They represent the possible values for \(p\) and \(q\).
Product-Sum Method
The Product-Sum Method is a technique for factoring quadratic equations. It involves finding two numbers that multiply to the constant term and add to the coefficient of the linear term. Given our quadratic equation:
\(x^2 + 4x - 5\),
we need two numbers that multiply to \(-5\) (the constant term) and add up to \(4\) (the coefficient of the linear term). These numbers are \(5\) and \(-1\). So, we rewrite the middle term using these numbers:
\(x^2 + 5x - x - 5\).
Next, group the terms to factor by grouping:
\(x(x + 5) - 1(x + 5)\).
We can then factor out the common binomial, giving us the final factors:
\((x + 5)(x - 1)\).
This method simplifies solving the quadratic equation by breaking it down into smaller, more manageable steps.
\(x^2 + 4x - 5\),
we need two numbers that multiply to \(-5\) (the constant term) and add up to \(4\) (the coefficient of the linear term). These numbers are \(5\) and \(-1\). So, we rewrite the middle term using these numbers:
\(x^2 + 5x - x - 5\).
Next, group the terms to factor by grouping:
\(x(x + 5) - 1(x + 5)\).
We can then factor out the common binomial, giving us the final factors:
\((x + 5)(x - 1)\).
This method simplifies solving the quadratic equation by breaking it down into smaller, more manageable steps.