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Magazine Chapter 10: Problem 57
Solve each equation. $$ x-6 x^{1 / 2}=-8 $$
Short Answer
Expert verified
Solutions are \( x = 16 \) and \( x = 4 \).
Step by step solution
01
Substitute the Square Root
Let's make a substitution to simplify the equation. Let \( y = x^{1/2} \). Then \( y^2 = x \). Substitute into the equation, to give \( y^2 - 6y = -8 \).
02
Rearrange the Equation
Rearrange the quadratic equation to standard form: \( y^2 - 6y + 8 = 0 \).
03
Factor the Quadratic Equation
Factor the quadratic equation \( y^2 - 6y + 8 = 0 \) as \( (y - 4)(y - 2) = 0 \).
04
Solve for y
Setting each factor equal to zero gives \( y - 4 = 0 \) or \( y - 2 = 0 \). So, \( y = 4 \) or \( y = 2 \).
05
Substitute Back to Find x
Recall that \( y = x^{1/2} \). So, for \( y = 4 \), we have \( x^{1/2} = 4 \) leading to \( x = 16 \). For \( y = 2 \), we have \( x^{1/2} = 2 \) leading to \( x = 4 \).
06
Verify Each Solution
Substitute \( x = 16 \) into the original equation to check: \( 16 - 6 \cdot 4 = 16 - 24 = -8 \). Substitute \( x = 4 \) into the original equation to check: \( 4 - 6 \cdot 2 = 4 - 12 = -8 \). Both values satisfy the original equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Quadratics
Factoring quadratics is a key method used to solve quadratic equations. A quadratic equation is typically in the form of \( ax^2 + bx + c = 0 \). To factor a quadratic, you find two binomials that multiply together to give you the original equation. Here is what you need to know:
- Identify the Standard Form: Make sure your equation is in the form of \( ax^2 + bx + c = 0 \). For example, with the equation \( y^2 - 6y + 8 = 0 \), the standard form is already present.
- Find the Factors: Look for two numbers that multiply to \( c \) (the constant term) and add to \( b \) (the coefficient of the \( y \) term). In the example \( y^2 - 6y + 8 = 0 \), we need two numbers that multiply to 8 and add to -6. These numbers are -4 and -2.
- Write as Binomials: Using the numbers you find, express the quadratic as a product of two binomials: \((y - 4)(y - 2) = 0\).
Solving Equations
Once a quadratic equation is factored, solving it involves setting each factor equal to zero. This is based on the zero-product property, which states that if the product of two factors is zero, at least one factor must be zero.
Here鈥檚 how it works:
Here鈥檚 how it works:
- Set Each Factor Equal to Zero: From the factored form \((y - 4)(y - 2) = 0\), set \(y - 4 = 0\) and \(y - 2 = 0\).
- Solve for the Variable: Solving each equation for \( y \):
\( y - 4 = 0 \) gives \( y = 4 \).
\( y - 2 = 0 \) gives \( y = 2 \).
Substitution Method
The substitution method is a strategic technique used to simplify complex equations and solve them more easily. It involves replacing a part of the equation with a simpler expression or variable, solving it, and then substituting back.
For example, in the equation \( x - 6x^{1/2} = -8 \), \( y = x^{1/2} \) was used as a substitution.
For example, in the equation \( x - 6x^{1/2} = -8 \), \( y = x^{1/2} \) was used as a substitution.
- Choose a Substitution: The first step is identifying a substitution that simplifies the equation. Here, letting \( y = x^{1/2} \) transforms \( x \) into \( y^2 \) because \( (x^{1/2})^2 = x \).
- Transform the Equation: Substitute these into the original equation to obtain \( y^2 - 6y = -8 \), which is noticeably simpler.
- Resolve and Back-substitute: After obtaining solutions for \( y \) using the quadratic method, substitute back for \( x \). For instance, if \( y = 4 \), then \( x = 4^2 = 16 \).
