Problem 24 Solve. \(\sqrt{10-3 q}-6=q\)... [FREE SOLUTION]

Chapter 11: Problem 24

Solve. \(\sqrt{10-3 q}-6=q\)

Short Answer

Expert verified

The given equation is \(\sqrt{10-3q}-6=q\). After isolating the square root, squaring both sides, rearranging, and simplifying, we get the quadratic equation \(0=q^2+15q+26\). Using the quadratic formula, we obtain two possible solutions, \(q=-2\) and \(q=-13\). However, upon checking both solutions, neither of them satisfy the original equation. Therefore, there are no valid solutions.

Step by step solution

Isolate the square root term

We want to have the square root term on one side of the equation, so we'll add 6 to both sides: \[\sqrt{10-3q}=q+6\]

Square both sides of the equation

Now, we'll square both sides to eliminate the square root: \[(\sqrt{10-3q})^2=(q+6)^2\] This simplifies to: \[10-3q=q^2+12q+36\]

Rearrange and simplify the equation

We want to have the equation in the quadratic form \(ax^2+bx+c=0\), so we'll move all terms to the left side of the equation: \[0=q^2+12q+36+3q-10\] Simplify: \[0=q^2+15q+26\]

Solve for \(q\)

Now, we'll solve the equation using the quadratic formula: \[q=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] In our equation, \(a=1\), \(b=15\), and \(c=26\). Plugging in the values, we get: \[q=\frac{-15\pm\sqrt{15^2-4(26)}}{2}\] Calculate the discriminant (\(b^2-4ac\)): \[15^2-4(26)=121\] Now, we can write and solve the equation for \(q\): \[q=\frac{-15\pm\sqrt{121}}{2}\] We have two possible solutions: \[q=\frac{-15+11}{2}=-2\] \[q=\frac{-15-11}{2}=-13\]

Check the solutions

Let's check both solutions to make sure they are valid: For \(q=-2\): \[\sqrt{10-3(-2)}-6\stackrel{?}{=} -2\] \[\sqrt{16}-6=2-6=-4\] This solution is invalid, because it doesn't satisfy the original equation. For \(q=-13\): \[\sqrt{10-3(-13)}-6\stackrel{?}{=} -13\] \[\sqrt{49}-6=7-6=1\] This solution is also invalid, because it doesn't satisfy the original equation. So, there are no valid solutions for the given equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Radical Equations

Solving radical equations can seem a bit tricky at first, but with practice, it becomes straightforward. A radical equation is one that features a variable inside a radical, usually a square root. Our goal is to isolate the radical term and eliminate it to solve the equation.

Here's a simplified way to approach it:

First, isolate the radical term on one side of the equation. This makes it easier to work with.
Next, square both sides of the equation. This step removes the square root, allowing you to handle the equation algebraically.
Finally, solve the equation like any other鈥攐ften using methods for solving quadratic equations. This process might introduce extraneous roots, so it's best to check your solutions against the original equation.

By following these steps, you can systematically solve even the most intimidating-looking radical equations. Don't forget that checking your solutions is crucial, as demonstrated in our original exercise where both potential solutions turned out to be invalid.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). This is especially useful when the equation doesn't factor easily.

The formula is represented as:

\( q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here's a simple breakdown of how to use it:

Identify the coefficients \(a\), \(b\), and \(c\) in your quadratic equation.
Substitute these coefficients into the formula.
Compute the discriminant, \(b^2 - 4ac\), which determines the nature of the roots.
Calculate the two potential solutions using the "plus or minus" component.

The quadratic formula is versatile, offering solutions when other methods might not work. It's always important, however, to verify these solutions to confirm if they make sense in the context of the original problem.

Discriminant

The discriminant is a key component of the quadratic formula and is crucial in understanding the nature of the equation's roots. It's calculated using \(b^2 - 4ac\). This value reveals important information about the solutions:

If the discriminant is positive, the equation has two distinct real roots.
If the discriminant is zero, there is exactly one real root, meaning the roots are repeated or the same.
If the discriminant is negative, the roots are complex (non-real), indicating the graph of the equation does not intersect the x-axis.

In our exercise, the discriminant was \(121\), which is positive. This initially suggested two possible real solutions. However, checking these solutions against the original radical equation revealed that neither worked, a common occurrence in solving radical equations due to possible extraneous solutions introduced during the squaring step.

91影视

Solve. \(\sqrt{10-3 q}-6=q\)

Short Answer

Step by step solution

Isolate the square root term

Square both sides of the equation

Rearrange and simplify the equation

Solve for \(q\)

Check the solutions

Key Concepts

Solving Radical Equations

Quadratic Formula

Discriminant

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