Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Certainly! Let's solve the given equation step by step.
Given equation:
[tex]\[ \frac{3}{2x^2} = \frac{1}{x} - \frac{1}{8} \][/tex]
Step 1: Eliminate fractions by finding a common denominator.
The least common multiple (LCM) of [tex]\( 2x^2 \)[/tex], [tex]\( x \)[/tex], and [tex]\( 8 \)[/tex] is [tex]\( 8x^2 \)[/tex].
Multiply both sides of the equation by [tex]\( 8x^2 \)[/tex] to clear the fractions:
[tex]\[ 8x^2 \cdot \frac{3}{2x^2} = 8x^2 \cdot \left( \frac{1}{x} - \frac{1}{8} \right) \][/tex]
Step 2: Simplify the equation after multiplying by [tex]\( 8x^2 \)[/tex]:
[tex]\[ 8x^2 \cdot \frac{3}{2x^2} = 8x^2 \cdot \frac{1}{x} - 8x^2 \cdot \frac{1}{8} \][/tex]
This reduces to:
[tex]\[ 4 \cdot 3 = 8x - x^2 \][/tex]
So, we get:
[tex]\[ 12 = 8x - x^2 \][/tex]
Step 3: Rearrange the equation to form a standard quadratic equation:
Move all terms to one side to set the equation to 0,
[tex]\[ x^2 - 8x + 12 = 0 \][/tex]
Step 4: Solve the quadratic equation using the quadratic formula:
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the equation [tex]\( x^2 - 8x + 12 = 0 \)[/tex], the coefficients are:
[tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 12 \)[/tex].
Substitute these values into the quadratic formula:
[tex]\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{64 - 48}}{2} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ x = \frac{8 \pm 4}{2} \][/tex]
So the solutions are:
[tex]\[ x = \frac{8 + 4}{2} = \frac{12}{2} = 6 \][/tex]
and
[tex]\[ x = \frac{8 - 4}{2} = \frac{4}{2} = 2 \][/tex]
Therefore, the solutions to the equation [tex]\( \frac{3}{2x^2} = \frac{1}{x} - \frac{1}{8} \)[/tex] are:
[tex]\[ \boxed{2 \text{ and } 6} \][/tex]
Given equation:
[tex]\[ \frac{3}{2x^2} = \frac{1}{x} - \frac{1}{8} \][/tex]
Step 1: Eliminate fractions by finding a common denominator.
The least common multiple (LCM) of [tex]\( 2x^2 \)[/tex], [tex]\( x \)[/tex], and [tex]\( 8 \)[/tex] is [tex]\( 8x^2 \)[/tex].
Multiply both sides of the equation by [tex]\( 8x^2 \)[/tex] to clear the fractions:
[tex]\[ 8x^2 \cdot \frac{3}{2x^2} = 8x^2 \cdot \left( \frac{1}{x} - \frac{1}{8} \right) \][/tex]
Step 2: Simplify the equation after multiplying by [tex]\( 8x^2 \)[/tex]:
[tex]\[ 8x^2 \cdot \frac{3}{2x^2} = 8x^2 \cdot \frac{1}{x} - 8x^2 \cdot \frac{1}{8} \][/tex]
This reduces to:
[tex]\[ 4 \cdot 3 = 8x - x^2 \][/tex]
So, we get:
[tex]\[ 12 = 8x - x^2 \][/tex]
Step 3: Rearrange the equation to form a standard quadratic equation:
Move all terms to one side to set the equation to 0,
[tex]\[ x^2 - 8x + 12 = 0 \][/tex]
Step 4: Solve the quadratic equation using the quadratic formula:
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the equation [tex]\( x^2 - 8x + 12 = 0 \)[/tex], the coefficients are:
[tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 12 \)[/tex].
Substitute these values into the quadratic formula:
[tex]\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{64 - 48}}{2} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ x = \frac{8 \pm 4}{2} \][/tex]
So the solutions are:
[tex]\[ x = \frac{8 + 4}{2} = \frac{12}{2} = 6 \][/tex]
and
[tex]\[ x = \frac{8 - 4}{2} = \frac{4}{2} = 2 \][/tex]
Therefore, the solutions to the equation [tex]\( \frac{3}{2x^2} = \frac{1}{x} - \frac{1}{8} \)[/tex] are:
[tex]\[ \boxed{2 \text{ and } 6} \][/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.