Problem 17
Question
Solve each equation in Exercises \(15-34\) by the square root property. $$ 5 x^{2}+1=51 $$
Step-by-Step Solution
Verified Answer
The solution to the equation \(5 x^{2}+1=51\) is \(x=±\sqrt{10}\).
1Step 1: Simplify the Equation
The given equation is \(5 x^{2}+1=51\). We first simplify this equation by subtracting 1 from both sides which renders it to be \(5 x^{2}=50\).
2Step 2: Isolate the Squared Variable
The next step is to isolate the squared variable \(x^{2}\) which can be done by dividing both sides of the equation by 5. This gives us \(x^{2}=10\).
3Step 3: Square root property
Now apply the square root property to both sides of the equation. Remember, whenever we take the square root of both sides in an equation, we should always consider both the positive and negative roots. Therefore, \(x=±\sqrt{10}\).
Key Concepts
Quadratic EquationSolving EquationsAlgebra Calculations
Quadratic Equation
A quadratic equation is an algebraic equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients with \(a eq 0\). The highest degree of the variable is 2, which is why it is called a quadratic equation. These equations can represent curves known as parabolas when graphed on a coordinate plane.
Quadratics can take various forms and not every quadratic equation needs to be complete with all three terms \(ax^2\), \(bx\), and \(c\). In the case of the problem given, only the \(ax^2\) term is present initially, simplifying the structure of the equation to \(5x^2 + 1 = 51\).
Quadratic equations have up to two solutions or roots. These solutions can be found using methods such as factoring, completing the square, or using the quadratic formula. However, in this particular instance, the equation involves a slightly adapted form known as the square root property.
Quadratics can take various forms and not every quadratic equation needs to be complete with all three terms \(ax^2\), \(bx\), and \(c\). In the case of the problem given, only the \(ax^2\) term is present initially, simplifying the structure of the equation to \(5x^2 + 1 = 51\).
Quadratic equations have up to two solutions or roots. These solutions can be found using methods such as factoring, completing the square, or using the quadratic formula. However, in this particular instance, the equation involves a slightly adapted form known as the square root property.
Solving Equations
Solving equations involves finding the value(s) of the variable that make the equation true. This means manipulating the equation until the variable is isolated on one side of the equation. Through systematic operations, the unknown becomes known.
The given equation \(5x^2 + 1 = 51\) is a quadratic equation that can be solved using the square root property. Solving begins with simplifying and transforming the equation:
This process of solving entails performing operations that maintain the balance of the equation while simplifying it.
The given equation \(5x^2 + 1 = 51\) is a quadratic equation that can be solved using the square root property. Solving begins with simplifying and transforming the equation:
- First, subtract 1 from both sides to get \(5x^2 = 50\).
- Next, divide every term by 5 to isolate \(x^2\), which gives \(x^2 = 10\).
This process of solving entails performing operations that maintain the balance of the equation while simplifying it.
Algebra Calculations
In algebra, calculations involve using arithmetic operations in combination with algebraic principles to solve equations or simplify expressions.
In the context of the square root property, after isolating \(x^2\), the algebra calculation involves applying the square root to both sides of the equation. Mathematically, this step is expressed as:
This demonstrates the elegance of algebra calculations, which involve systematically applying rules and properties to uncover the desired solutions.
In the context of the square root property, after isolating \(x^2\), the algebra calculation involves applying the square root to both sides of the equation. Mathematically, this step is expressed as:
- Taking the square root of \(x^2 = 10\), we derive \(x = \pm\sqrt{10}\).
This demonstrates the elegance of algebra calculations, which involve systematically applying rules and properties to uncover the desired solutions.
Other exercises in this chapter
Problem 17
In Exercises \(9-20,\) find each product and write the result in standard form. $$(-5+i)(-5-i)$$
View solution Problem 17
Find all values of \(x\) satisfying the given conditions. \(y_{1}=\frac{1}{x}, y_{2}=\frac{1}{2 x}, y_{3}=\frac{1}{x-1},\) and the sum of 3 times \(y_{1}\) and
View solution Problem 17
Check all proposed solutions. $$ x-\sqrt{2 x+5}=5 $$
View solution Problem 17
Contain linear equations with constants in denominators. Solve equation. \(\frac{x}{3}=\frac{x}{2}-2\)
View solution