Problem 42 - Algebra 1: Concepts and Skills

Question

Solve the equation or write no real solution. Write the solutions as integers if possible. Otherwise, write them as radical expressions. $$ 7 x^{2}+30=9 $$

Step-by-Step Solution

Answer

The equation has no real solution.

Step 1: Isolate the \(x^2\) term

First, let's isolate the \(x^2\) term. We can do this by subtracting 30 from both sides of the equation. So, the equation \(7x^2 + 30 = 9\) simplifies to \(7x^2 = 9 - 30\) which simplifies further to \(7x^2 = -21\).

Step 2: Solve for \(x^2\)

Next, solve for \(x^2\). Do this by dividing both sides by 7. So, \(x^2 = -21/7\) simplifies to \(x^2 = -3\).

Step 3: Solve for \(x\)

Finally, compute for \(x\). The square root of a negative number is undefined in the real number system, so the equation has no real solution.

Key Concepts

Understanding Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and at least one arithmetic operation. In our exercise, we have the algebraic expression 7x^2 + 30. To solve for x, we need to perform operations that will simplify this expression and isolate x. The first step in handling such expressions typically involves gathering like terms and performing any required operations to get the variable of interest, in this case x^2, by itself.

To start, we subtract 30 from both sides to isolate the 7x^2 term. In general, keep the following tips in mind when dealing with algebraic expressions:

• Always combine like terms where possible.
• Use inverse operations to move terms from one side of the equation to the other. For instance, subtract if you have an addition or divide if you have a multiplication.
• Remember the order of operations (PEMDAS/BODMAS) when simplifying expressions.

Once the terms are isolated, we can move to the next step which often involves dealing with radical expressions or solving for the variable outright.

Working with Radical Expressions

Radical expressions involve roots, like square roots or cube roots. In our exercise, we encounter a radical expression after isolating x^2 when we attempt to solve for x. To find x, we need to take the square root of x^2, but there is a complication because x^2 equals a negative number, -3.

Remember that the square root of a negative number is not a real number—it's an imaginary number. This is because no real number squared will give a negative result. When dealing with radical expressions, try the following strategies:

• Look for opportunities to simplify the expression under the root, if it's a positive number.
• If you have a negative under the square root, acknowledge that there is no real solution in the set of real numbers.
• Be prepared to express roots of perfect squares as integers when possible.

In mathematics, especially algebra, working with radical expressions is key to solving equations that involve roots. The journey from recognizing that we have a negative under the square root to concluding there is no real solution is a quintessential aspect of understanding radicals.

Identifying No Real Solution

When we arrive at an equation where we have a negative number under a square root, such as x^2 = -3, we understand that this signifies there is 'no real solution' in the context of real numbers. In our quadratic equation example, after isolating the square term and simplifying, we end up with a negative number, which points to the fact that x^2 cannot equal a negative value when we're working within the real number system.

In case we are required to find a solution, we would have to extend our number system to include complex numbers, where the square root of -1 is defined as i, the imaginary unit. Since our initial instruction did not accommodate for complex solutions, we conclude with 'no real solution'.

It is vital to grasp that the absence of a real solution does not imply an error in the process, but rather the nature of square roots and the limitations within the set of real numbers. In higher-level mathematics, complex numbers are used to provide solutions to these types of equations, but that's beyond the scope of this particular problem.