Answer :
Certainly! Let's solve the equation step by step:
Given the equation:
[tex]\[ 4x^2 + 16 = -32 \][/tex]
1. Isolate the quadratic term: First, let's move the constant term on the left side to the right side by subtracting 16 from both sides of the equation:
[tex]\[ 4x^2 + 16 - 16 = -32 - 16 \][/tex]
[tex]\[ 4x^2 = -48 \][/tex]
2. Simplify: Now, we divide both sides by 4 to simplify the coefficient of \( x^2 \):
[tex]\[ x^2 = \frac{-48}{4} \][/tex]
[tex]\[ x^2 = -12 \][/tex]
3. Solve for \( x \): To find \( x \), we take the square root of both sides. Remember that taking the square root of a negative number involves imaginary numbers. Thus,
[tex]\[ x = \pm \sqrt{-12} \][/tex]
[tex]\[ x = \pm \sqrt{-1 \cdot 12} \][/tex]
[tex]\[ x = \pm \sqrt{-1} \cdot \sqrt{12} \][/tex]
Since the square root of \(-1\) is \(i\) (the imaginary unit), we get:
[tex]\[ x = \pm i \cdot \sqrt{12} \][/tex]
4. Simplify the square root of 12: The number 12 can be simplified further:
[tex]\[ \sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2 \cdot \sqrt{3} \][/tex]
Therefore,
[tex]\[ x = \pm i \cdot 2\sqrt{3} \][/tex]
So, the two solutions are:
[tex]\[ x_1 = -2i\sqrt{3} \][/tex]
[tex]\[ x_2 = 2i\sqrt{3} \][/tex]
In conclusion, the solutions to the equation \( 4x^2 + 16 = -32 \) are:
[tex]\[ x_1 = -2i\sqrt{3} \][/tex]
[tex]\[ x_2 = 2i\sqrt{3} \][/tex]
Given the equation:
[tex]\[ 4x^2 + 16 = -32 \][/tex]
1. Isolate the quadratic term: First, let's move the constant term on the left side to the right side by subtracting 16 from both sides of the equation:
[tex]\[ 4x^2 + 16 - 16 = -32 - 16 \][/tex]
[tex]\[ 4x^2 = -48 \][/tex]
2. Simplify: Now, we divide both sides by 4 to simplify the coefficient of \( x^2 \):
[tex]\[ x^2 = \frac{-48}{4} \][/tex]
[tex]\[ x^2 = -12 \][/tex]
3. Solve for \( x \): To find \( x \), we take the square root of both sides. Remember that taking the square root of a negative number involves imaginary numbers. Thus,
[tex]\[ x = \pm \sqrt{-12} \][/tex]
[tex]\[ x = \pm \sqrt{-1 \cdot 12} \][/tex]
[tex]\[ x = \pm \sqrt{-1} \cdot \sqrt{12} \][/tex]
Since the square root of \(-1\) is \(i\) (the imaginary unit), we get:
[tex]\[ x = \pm i \cdot \sqrt{12} \][/tex]
4. Simplify the square root of 12: The number 12 can be simplified further:
[tex]\[ \sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2 \cdot \sqrt{3} \][/tex]
Therefore,
[tex]\[ x = \pm i \cdot 2\sqrt{3} \][/tex]
So, the two solutions are:
[tex]\[ x_1 = -2i\sqrt{3} \][/tex]
[tex]\[ x_2 = 2i\sqrt{3} \][/tex]
In conclusion, the solutions to the equation \( 4x^2 + 16 = -32 \) are:
[tex]\[ x_1 = -2i\sqrt{3} \][/tex]
[tex]\[ x_2 = 2i\sqrt{3} \][/tex]
