Problem 45
Question
Solve each equation by making an appropriate substitution. $$x-13 \sqrt{x}+40-0$$
Step-by-Step Solution
Verified Answer
The solutions for the equation are \(x = 64\) and \(x = 25\).
1Step 1: Apply substitution
Substitute \(\sqrt{x}\) with \(u\). The equation will then be \(u^2 - 13u + 40 = 0\) where \(u = \sqrt{x}\).
2Step 2: Solve the quadratic equation
The quadratic equation can be factored into \((u - 8)(u - 5) = 0\). Setting each factor equal to zero gives the roots of the equation as \(u = 8\) and \(u = 5\).
3Step 3: Substitute u back in terms of x
Since \(u = \sqrt{x}\), we can substitute \(u\) back in terms of \(x\). This yields two equations which are \( \sqrt{x} = 8\) and \( \sqrt{x} = 5\). Squaring both sides of each equation to solve for \(x\) gives \(x = 64\) and \(x = 25\).
Key Concepts
Algebraic SubstitutionFactoring QuadraticsSquare Root Equations
Algebraic Substitution
Algebraic substitution is a powerful technique used to simplify certain types of equations or expressions. In the context of solving quadratic equations, this method can make a complex problem more tangible. For example, when confronted with an equation that contains a square root, like \(x-13\sqrt{x}+40=0\), it can be tricky to find a solution directly.
By introducing a new variable, say \(u\), to represent the square root of \(x\), so that \(u = \sqrt{x}\), we transform the original equation into a more familiar quadratic form, \(u^2 - 13u + 40 = 0\). This step effectively reduces the problem to one where standard quadratic solution methods can be applied. The substitution not only simplifies the calculation but also prevents potential errors that might arise from handling square root terms throughout the solution process.
By introducing a new variable, say \(u\), to represent the square root of \(x\), so that \(u = \sqrt{x}\), we transform the original equation into a more familiar quadratic form, \(u^2 - 13u + 40 = 0\). This step effectively reduces the problem to one where standard quadratic solution methods can be applied. The substitution not only simplifies the calculation but also prevents potential errors that might arise from handling square root terms throughout the solution process.
Factoring Quadratics
Factoring quadratics is a quintessential skill in algebra. When a quadratic equation is presented, one of the first methods attempted is usually factoring. This technique involves expressing the quadratic as a product of two binomials. The equation \(u^2 - 13u + 40 = 0\), derived from the original problem through substitution, is a prime candidate for factoring.
The process of factoring depends heavily on finding two numbers that not only multiply to give the constant term, in this case, 40, but also add up to the coefficient of the linear term, which is -13. Through a series of trial and error or insights into number pairs, we discover that 8 and 5 are the numbers that satisfy both conditions. Consequently, we can write the quadratic equation as \((u - 8)(u - 5) = 0\). This form makes it evident that the values of \(u\) are 8 and 5, which lead us one step closer to determining the value of \(x\).
The process of factoring depends heavily on finding two numbers that not only multiply to give the constant term, in this case, 40, but also add up to the coefficient of the linear term, which is -13. Through a series of trial and error or insights into number pairs, we discover that 8 and 5 are the numbers that satisfy both conditions. Consequently, we can write the quadratic equation as \((u - 8)(u - 5) = 0\). This form makes it evident that the values of \(u\) are 8 and 5, which lead us one step closer to determining the value of \(x\).
Square Root Equations
Solving square root equations requires a delicate balance of algebraic manipulation and care to ensure that the solutions obtained are valid. Once algebraic substitution and factoring quadratics lead us to the roots \(u = 8\) and \(u = 5\), we must revert back to the original variable \(x\) by undoing the substitution. This entails squaring both sides of the equations \(\sqrt{x} = 8\) and \(\sqrt{x} = 5\) to eliminate the square root.
By squaring, the equations transform into \(x = 64\) and \(x = 25\), respectively. It's crucial to verify that these solutions indeed satisfy the original equation. Since squaring is an irreversible process—it affects both the positive and negative roots—it might introduce extraneous solutions. However, since square roots are inherently non-negative, only the positive solutions are considered valid. Techniques used to solve square root equations help demystify more complicated problems and tie together an understanding of how algebraic operations affect an equation's solutions.
By squaring, the equations transform into \(x = 64\) and \(x = 25\), respectively. It's crucial to verify that these solutions indeed satisfy the original equation. Since squaring is an irreversible process—it affects both the positive and negative roots—it might introduce extraneous solutions. However, since square roots are inherently non-negative, only the positive solutions are considered valid. Techniques used to solve square root equations help demystify more complicated problems and tie together an understanding of how algebraic operations affect an equation's solutions.
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