Solve For Q: 4q² - 63 = 0

When we encounter a mathematical problem that asks us to solve for q in an equation like $4q^2 - 63 = 0$, we're essentially being asked to find the value or values of the variable q that make the equation true. This particular equation is a type of quadratic equation, distinguished by the presence of a variable raised to the second power ($q^2$). These equations often appear in algebra and have a wide range of applications in fields such as physics, engineering, and economics. The process of solving them typically involves isolating the variable term and then taking the square root. We'll walk through the steps to find the solutions for q in this specific equation, ensuring our answer is in a simplified, rationalized form as requested.

To begin solving for q in the equation $4q^2 - 63 = 0$, our primary goal is to isolate the $q^2$ term. This involves a series of algebraic manipulations. First, we want to move the constant term, -63, to the other side of the equation. We can achieve this by adding 63 to both sides of the equation. This operation maintains the equality of the equation. So, we have $4q^2 - 63 + 63 = 0 + 63$, which simplifies to $4q^2 = 63$. Now, the $q^2$ term is closer to being isolated. The next step is to get rid of the coefficient 4 that is multiplying $q^2$. To do this, we divide both sides of the equation by 4. This gives us $\frac{4q^2}{4} = \frac{63}{4}$. After this division, we are left with $q^2 = \frac{63}{4}$. At this stage, we have successfully isolated $q^2$. The equation now tells us that $q$ squared is equal to $\frac{63}{4}$. This is a crucial step because it sets up the final operation needed to find the values of q.

Having reached the step where $q^2 = \frac{63}{4}$, the next logical step to solve for q is to take the square root of both sides of the equation. Remember that when we take the square root of a number, there are always two possible results: a positive and a negative root. This is because both a positive number squared and its negative counterpart squared yield the same positive result. Therefore, when we take the square root of $q^2$, we get $q$, and when we take the square root of $\frac{63}{4}$, we must account for both the positive and negative possibilities. So, we write $q =

\pm\sqrt{\frac{63}{4}}$. The square root of a fraction is equal to the square root of the numerator divided by the square root of the denominator, meaning $\sqrt{\frac{63}{4}} = \frac{\sqrt{63}}{\sqrt{4}}$. We know that $\sqrt{4}$ is 2, so the expression simplifies to $q =

\pm\frac{\sqrt{63}}{2}$. Now, we need to consider if $\sqrt{63}$ can be simplified. We look for perfect square factors within 63. The number 63 can be factored as $9 \times 7$. Since 9 is a perfect square ($3^2$), we can rewrite $\sqrt{63}$ as $\sqrt{9 \times 7}$. Using the property of square roots that $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$, we get $\sqrt{9} \times \sqrt{7}$, which is $3\sqrt{7}$. Substituting this back into our equation for q, we get $q =

\pm\frac3\sqrt{7}}{2}$. This gives us our two solutions for q in their simplified and rationalized form $q = \frac{3\sqrt{7}{2}$ and $q = -\frac{3\sqrt{7}}{2}$. The term 'rationalized' in this context means that there is no square root in the denominator of our fraction, which is already the case here.

In conclusion, when faced with the task to solve for q in the quadratic equation $4q^2 - 63 = 0$, we followed a systematic approach. We first isolated the $q^2$ term by adding 63 to both sides, resulting in $4q^2 = 63$, and then dividing by 4 to get $q^2 = \frac{63}{4}$. The crucial next step was to take the square root of both sides, remembering to include both positive and negative roots, leading to $q =

\pm\sqrt{\frac{63}{4}}$. We then simplified the square root of the fraction by simplifying the square root of the numerator and the denominator separately. Specifically, $\sqrt{63}$ was simplified to $3\sqrt{7}$ because $63 = 9

imes

7$ and $\sqrt{9}=3$, while $\sqrt{4}$ simplified to 2. This resulted in the final simplified and rationalized solutions for q as $q = \frac{3\sqrt{7}}{2}$ and $q = -\frac{3\sqrt{7}}{2}$. These are the two values that, when substituted back into the original equation $4q^2 - 63 = 0$, will make the statement true. This process is fundamental in algebra and is a stepping stone to understanding more complex equations and their applications in various scientific and engineering disciplines.

For further exploration into solving quadratic equations and understanding their mathematical properties, you can visit Khan Academy's Quadratic Equations section. This resource offers comprehensive lessons, practice exercises, and detailed explanations that can deepen your understanding of this essential mathematical concept.