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Chapter 9: Problem 11

\(4 x^{2}=100\)

Short Answer

Expert verified
x = 5 or x = -5

Step by step solution

01

Isolate the Quadratic Term

Start by dividing both sides of the equation by 4 to isolate the quadratic term. This gives: \[ \frac{4x^2}{4} = \frac{100}{4} \]which simplifies to:\[ x^2 = 25 \]
02

Solve for x

Next, solve for x by taking the square root of both sides of the equation:\[ x = \pm \sqrt{25} \]which simplifies to:\[ x = \pm 5 \]
03

State the Final Answer

The solutions to the equation are:\[ x = 5 \]and\[ x = -5 \]

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Key Concepts

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

Isolate the Quadratic Term
To solve a quadratic equation like \(4 x^{2}=100\), we first need to isolate the quadratic term, which is \(4x^2\). Because this term is multiplied by 4, we divide both sides of the equation by 4. This step simplifies the equation and gets rid of the coefficient in front of the \(x^2\) term.
For example, dividing both sides by 4: \[ \frac{4x^2}{4} = \frac{100}{4} \] simplifies to \[ x^2 = 25 \]
So now the quadratic term \(x^2\) is isolated, making it easier to solve in the next steps.
Square Root Method
After isolating the quadratic term, the next step is to solve for x using the square root method. In our equation \( x^2 = 25 \), the goal is to find x. To do this, we take the square root of both sides of the equation.
This gives us: \[ x = \pm \sqrt{25} \]
It鈥檚 important to include both the positive and negative solutions because both \(5\) and \(-5\) squared will give us 25.
This step simplifies our equation to: \[ x = \pm 5 \].
Positive and Negative Solutions
One essential concept in solving quadratic equations is to recognize that taking the square root of a number produces two solutions: one positive and one negative. In our example, when we calculated \( x = \pm \sqrt{25} \), we got two possible solutions:
\[ x = 5 \]
\[ x = -5 \]
These solutions mean that both \(5\) and \(-5\) are roots of the original equation \(4x^2 = 100\).
Therefore, always remember that when solving \( x^2 = a \), there are two solutions, \(+\sqrt{a}\) and \(-\sqrt{a}.......\).

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