Problem 43
Question
For the following problems, solve the equations using the quadratic formula. $$ x^{2}+4 x=2 x-5 $$
Step-by-Step Solution
Verified Answer
Answer: The two solutions for y are 5 and -5.
1Step 1: Divide both sides by -3
We can start by dividing both sides of the equation by -3 to isolate the y² term:
$$
\frac{-3y^2}{-3} = \frac{-75}{-3}
$$
$$
y^2 = 25
$$
2Step 2: Take the square root of both sides
Now, take the square root of both sides to find the value of y:
$$
\sqrt{y^2} = \pm \sqrt{25}
$$
$$
y = \pm 5
$$
So the two solutions for y are 5 and -5.
Key Concepts
Solving Quadratic EquationsSquare RootIsolate VariableAlgebraic Operations
Solving Quadratic Equations
Quadratic equations are mathematical expressions that involve a squared term. They have the general form of:
\( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the unknown variable we want to solve for. In the above exercise, where \( -3y^2 = -75 \), we're dealing with a simpler form of a quadratic equation, with \( b \) and \( c \) essentially being equal to zero.
The solution to a quadratic equation can be found through various methods, one of which is extraction of roots, also known as the 'square root method'. This method is particularly effective when the quadratic equation can be simplified to the form \( y^2 = \text{constant} \). The process involves isolating the squared term and finding its square root to solve for the variable. The exercise provided is a perfect example of when to use extraction of roots, as we reduced the equation to \( y^2 = 25 \). It is important to note that squaring is a reversible operation, so taking the square root of both sides gives us the potential for two solutions: \( y = \text{±}\text{square root of constant} \). Hence, we find the two solutions for \( y \), which are 5 and -5.
\( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the unknown variable we want to solve for. In the above exercise, where \( -3y^2 = -75 \), we're dealing with a simpler form of a quadratic equation, with \( b \) and \( c \) essentially being equal to zero.
The solution to a quadratic equation can be found through various methods, one of which is extraction of roots, also known as the 'square root method'. This method is particularly effective when the quadratic equation can be simplified to the form \( y^2 = \text{constant} \). The process involves isolating the squared term and finding its square root to solve for the variable. The exercise provided is a perfect example of when to use extraction of roots, as we reduced the equation to \( y^2 = 25 \). It is important to note that squaring is a reversible operation, so taking the square root of both sides gives us the potential for two solutions: \( y = \text{±}\text{square root of constant} \). Hence, we find the two solutions for \( y \), which are 5 and -5.
Square Root
The square root of a number is a value that, when multiplied by itself, gives the original number. The symbol for square root is \( \sqrt{} \). For example, since \( 5 \times 5 = 25 \), the square root of 25 is 5. This is represented as \( \sqrt{25} = 5 \).
The concept of square roots is pivotal when solving quadratic equations by extraction of roots. When we encounter an equation like \( y^2 = 25 \), we can deduce that \( y \) could be either the positive or negative value whose square gives 25, which is why we express the solution as \( y = \text{±}5 \), taken from \( y = \text{±} \sqrt{25} \). It's important to remember that every positive number has two square roots: a positive and a negative root. However, there is no real square root for negative numbers unless we involve complex numbers, which are beyond the scope of this article.
The concept of square roots is pivotal when solving quadratic equations by extraction of roots. When we encounter an equation like \( y^2 = 25 \), we can deduce that \( y \) could be either the positive or negative value whose square gives 25, which is why we express the solution as \( y = \text{±}5 \), taken from \( y = \text{±} \sqrt{25} \). It's important to remember that every positive number has two square roots: a positive and a negative root. However, there is no real square root for negative numbers unless we involve complex numbers, which are beyond the scope of this article.
Isolate Variable
Isolating the variable in an equation means rearranging the equation so that the variable we're solving for is on one side of the equation by itself. It's a fundamental step in algebra, crucial for solving any equation. The process usually involves several algebraic operations: addition, subtraction, multiplication, division, and sometimes taking roots or applying exponents.
In our example, we isolated \( y^2 \) by dividing both sides by -3. This simplified our equation to \( y^2 = 25 \), where \( y^2 \) is now isolated on one side. Isolating the variable allows us to perform operations such as taking the square root without affecting other terms in the equation, leading us to the correct value or values for the variable.
In our example, we isolated \( y^2 \) by dividing both sides by -3. This simplified our equation to \( y^2 = 25 \), where \( y^2 \) is now isolated on one side. Isolating the variable allows us to perform operations such as taking the square root without affecting other terms in the equation, leading us to the correct value or values for the variable.
Algebraic Operations
Algebraic operations refer to the set of mathematical procedures used to manipulate expressions and solve equations. The basics include addition, subtraction, multiplication, and division. These operations follow specific properties and rules that govern how we can re-arrange and simplify equations. In the context of solving quadratic equations, we might add or subtract terms to move them across the equation's sides or multiply and divide to simplify terms.
For instance, in solving \( -3y^2 = -75 \), we first divided by -3 to simplify the quadratic term to \( y^2 \). This division is an algebraic operation that helped to prepare the equation for the extraction of roots. Mastery of algebraic operations is essential for effectively isolating variables and solving any algebraic equation.
For instance, in solving \( -3y^2 = -75 \), we first divided by -3 to simplify the quadratic term to \( y^2 \). This division is an algebraic operation that helped to prepare the equation for the extraction of roots. Mastery of algebraic operations is essential for effectively isolating variables and solving any algebraic equation.
Other exercises in this chapter
Problem 42
For the following problems, solve for the indicated variable. $$ k^{2}=p^{2} q^{2} r^{2}, \text { for } k $$
View solution Problem 42
For the following problems, use the zero-factor property to solve the equations. $$ 9(a-4)=0 $$
View solution Problem 43
Use the method of extraction of roots to solve \((x-2)^{2}=25\).
View solution Problem 43
For the following problems, solve the equations, if possible. $$ y^{2}-1=0 $$
View solution