Problem 72
Question
Solve using the square root property. Simplify all radicals. $$ (2 x-5)^{2}-180=0 $$
Step-by-Step Solution
Verified Answer
The solutions are \( x = 3\sqrt{5} + \frac{5}{2} \) and \( x = -3\sqrt{5} + \frac{5}{2} \).
1Step 1 - Isolate the squared term
Start by isolating the squared term \( (2x - 5)^2 \): \ (2 x - 5)^{2} - 180 = 0 \ Add 180 to both sides: \ (2 x - 5)^{2} = 180
2Step 2 - Apply the square root property
Take the square root of both sides of the equation: \ \( \sqrt{(2x - 5)^2} = \sqrt{180} \) \ Simplify the equation: \ \( 2x - 5 = \pm \sqrt{180} \)
3Step 3 - Simplify the radical
Simplify \(\sqrt{180}\): \ \( \sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5} \) \ Substitute this back into the equation: \ \( 2x - 5 = \pm 6\sqrt{5} \)
4Step 4 - Solve for x
Separate into two equations: \ \( 2x - 5 = 6\sqrt{5} \) and \( 2x - 5 = -6\sqrt{5} \) \ Solve each: \ For \( 2x - 5 = 6\sqrt{5} \): \ Add 5 to both sides: \ \( 2x = 6\sqrt{5} + 5 \) \ Divide by 2: \ \( x = 3\sqrt{5} + \frac{5}{2} \) \ For \( 2x - 5 = -6\sqrt{5} \): \ Add 5 to both sides: \ \( 2x = -6\sqrt{5} + 5 \) \ Divide by 2: \ \( x = -3\sqrt{5} + \frac{5}{2} \)
Key Concepts
square root propertyisolating squared termssimplifying radicals
square root property
One useful tool in solving quadratic equations is the square root property.
This property states that if you have an equation of the form \( k^2 = c \), then \( k = \pm\sqrt{c} \).
In simpler terms, when you have a squared term alone on one side of the equation, you can take the square root of both sides to solve for the variable inside the squared term.
This is highly useful because it can simplify otherwise complex algebraic problems into simpler ones.
Consider the equation \( (2x - 5)^2 - 180 = 0 \).
Our squared term here is \((2x - 5)^2\).
First, we isolate this term by adding 180 to both sides.
This leaves us with \( (2x - 5)^2 = 180 \).
Now, applying the square root property, we take the square root of both sides:
\[ \sqrt{(2x - 5)^2} = \sqrt{180} \]
Simplifying gives us:
\[ 2x - 5 = \pm\sqrt{180} \]
This property states that if you have an equation of the form \( k^2 = c \), then \( k = \pm\sqrt{c} \).
In simpler terms, when you have a squared term alone on one side of the equation, you can take the square root of both sides to solve for the variable inside the squared term.
This is highly useful because it can simplify otherwise complex algebraic problems into simpler ones.
Consider the equation \( (2x - 5)^2 - 180 = 0 \).
Our squared term here is \((2x - 5)^2\).
First, we isolate this term by adding 180 to both sides.
This leaves us with \( (2x - 5)^2 = 180 \).
Now, applying the square root property, we take the square root of both sides:
\[ \sqrt{(2x - 5)^2} = \sqrt{180} \]
Simplifying gives us:
\[ 2x - 5 = \pm\sqrt{180} \]
isolating squared terms
Isolating squared terms is a crucial step when solving quadratic equations using the square root property.
The goal is to get the squared term by itself on one side of the equation.
Let's revisit our original equation, \( (2x - 5)^2 - 180 = 0 \).
To isolate the squared term \( (2x - 5)^2 \), we need to eliminate the other terms around it.
Start by adding 180 to both sides:
\[ (2x - 5)^2 - 180 + 180 = 0 + 180 \]
This simplifies to:
\[ (2x - 5)^2 = 180 \]
Now the squared term is isolated, and we can use further methods to solve for the variable.
The goal is to get the squared term by itself on one side of the equation.
Let's revisit our original equation, \( (2x - 5)^2 - 180 = 0 \).
To isolate the squared term \( (2x - 5)^2 \), we need to eliminate the other terms around it.
Start by adding 180 to both sides:
\[ (2x - 5)^2 - 180 + 180 = 0 + 180 \]
This simplifies to:
\[ (2x - 5)^2 = 180 \]
Now the squared term is isolated, and we can use further methods to solve for the variable.
simplifying radicals
Simplifying radicals is the process of breaking down a complex square root into simpler, more manageable terms.
When you take the square root of a number, it helps to look for perfect squares inside the number.
For example, in our exercise, we have \( \sqrt{180} \).
Notice that 180 can be factored into 36 and 5, where 36 is a perfect square:
\[ 180 = 36 \times 5 \]
Therefore, \( \sqrt{180} \) becomes:
\[ \sqrt{36 \times 5} = \sqrt{36} \times \sqrt{5} = 6\sqrt{5} \]
This simplification helps us rewrite the square root in its simplest form.
So, our equation \( 2x - 5 = \pm\sqrt{180} \) changes to:
\( 2x - 5 = \pm6\sqrt{5} \).
From here, we can solve for x by further isolating x and dividing both sides by 2.
These steps lead us to our final solution.
When you take the square root of a number, it helps to look for perfect squares inside the number.
For example, in our exercise, we have \( \sqrt{180} \).
Notice that 180 can be factored into 36 and 5, where 36 is a perfect square:
\[ 180 = 36 \times 5 \]
Therefore, \( \sqrt{180} \) becomes:
\[ \sqrt{36 \times 5} = \sqrt{36} \times \sqrt{5} = 6\sqrt{5} \]
This simplification helps us rewrite the square root in its simplest form.
So, our equation \( 2x - 5 = \pm\sqrt{180} \) changes to:
\( 2x - 5 = \pm6\sqrt{5} \).
From here, we can solve for x by further isolating x and dividing both sides by 2.
These steps lead us to our final solution.
Other exercises in this chapter
Problem 71
Solve using the square root property. Simplify all radicals. $$ (4 x-1)^{2}-48=0 $$
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Solve for \(x .\) Assume that a and \(b\) represent positive real numbers. \(9 x^{2}-25 a=0\)
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The following exercises are not grouped by type. Solve each equation. (Exercises 83 and 84 require knowledge of complex numbers.) \(\left(x-\frac{1}{2}\right)^{
View solution Problem 73
Solve for \(x .\) Assume that a and \(b\) represent positive real numbers. \((5 x-2 b)^{2}=3 a\)
View solution