Solve the exercise:
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Solve the exercise:
To solve this equation, we'll follow these steps:
Now, let's work through each step:
Step 1: We begin with the equation . Our objective is to find the value(s) of that satisfy this equation. The equation is already composed of square terms and constants, providing a clue to equate similar terms and solve for .
Step 2: Let's rearrange the given equation:
\begin{align*}
x^2 + 64 &= 3x^2 + 16
\end{align*}
Subtract from both sides:
\begin{align*}
64 &= 3x^2 - x^2 + 16
\end{align*}
Which simplifies to:
\begin{align*}
64 &= 2x^2 + 16
\end{align*}
Next, subtract 16 from both sides:
\begin{align*}
64 - 16 &= 2x^2 \\
48 &= 2x^2
\end{align*}
Now, divide both sides by 2 to further isolate :
\begin{align*}
24 &= x^2
\end{align*}
Step 3: Solve for by taking the square root of both sides. Remember to consider both the positive and negative roots because squaring a number always yields a positive result:
\begin{align*}
x &= \pm \sqrt{24}
\end{align*}
Conclusion: The solutions to the equation are .
Therefore, the correct answer choice is: .
Solve the following equation:
\( 2x^2-8=x^2+4 \)
Because when you square any number, the result is always positive! Both and . So both values satisfy the original equation.
Yes! . So the final answer could also be written as ±2√6, but ±√24 is equally correct.
Move all x² terms to one side and all constants to the other. This gives you the standard form x² = (number), making it easy to solve by taking square roots.
Then the equation has no real solutions! You cannot take the square root of a negative number in real numbers. Always check your arithmetic if this happens.
Always expand numerical powers immediately! makes the equation much clearer: .
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