Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Sure, let's solve these equations step-by-step.
### 1) [tex]\( 16x - \frac{10}{x} = 27 \)[/tex]
First, multiply both sides of the equation by [tex]\( x \)[/tex] to get rid of the fraction:
[tex]\[ 16x^2 - 10 = 27x \][/tex]
Rearrange the equation to form a standard quadratic equation:
[tex]\[ 16x^2 - 27x - 10 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 16 \)[/tex], [tex]\( b = -27 \)[/tex], and [tex]\( c = -10 \)[/tex]:
[tex]\[ x = \frac{27 \pm \sqrt{(-27)^2 - 4 \cdot 16 \cdot (-10)}}{2 \cdot 16} \][/tex]
[tex]\[ x = \frac{27 \pm \sqrt{729 + 640}}{32} \][/tex]
[tex]\[ x = \frac{27 \pm \sqrt{1369}}{32} \][/tex]
[tex]\[ x = \frac{27 \pm 37}{32} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{64}{32} = 2 \][/tex]
[tex]\[ x = \frac{-10}{32} = -\frac{5}{16} \][/tex]
So, the solutions to the first equation are:
[tex]\[ x = 2 \quad \text{or} \quad x = -\frac{5}{16} \][/tex]
### 2) [tex]\( 9x^2 - 3x - 2 = 0 \)[/tex]
This is already a quadratic equation in standard form. Let's use the quadratic formula again: [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 9 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -2 \)[/tex]:
[tex]\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 9 \cdot (-2)}}{2 \cdot 9} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{9 + 72}}{18} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{81}}{18} \][/tex]
[tex]\[ x = \frac{3 \pm 9}{18} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{12}{18} = \frac{2}{3} \][/tex]
[tex]\[ x = \frac{-6}{18} = -\frac{1}{3} \][/tex]
So, the solutions to the second equation are:
[tex]\[ x = \frac{2}{3} \quad \text{or} \quad x = -\frac{1}{3} \][/tex]
### 3) [tex]\( \frac{1}{x-3} + \frac{2}{x-2} = \frac{8}{x} \)[/tex]
To solve this, first find a common denominator:
[tex]\[ \frac{(x-2) + 2(x-3)}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
[tex]\[ \frac{x - 2 + 2x - 6}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
[tex]\[ \frac{3x - 8}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
Cross-multiply to clear the fractions:
[tex]\[ x(3x - 8) = 8(x-3)(x-2) \][/tex]
[tex]\[ 3x^2 - 8x = 8(x^2 - 5x + 6) \][/tex]
[tex]\[ 3x^2 - 8x = 8x^2 - 40x + 48 \][/tex]
Rearrange the equation:
[tex]\[ 3x^2 - 8x - 8x^2 + 40x - 48 = 0 \][/tex]
[tex]\[ -5x^2 + 32x - 48 = 0 \][/tex]
Solve using the quadratic formula: [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -5 \)[/tex], [tex]\( b = 32 \)[/tex], and [tex]\( c = -48 \)[/tex]:
[tex]\[ x = \frac{-32 \pm \sqrt{32^2 - 4 \cdot (-5) \cdot (-48)}}{2 \cdot (-5)} \][/tex]
[tex]\[ x = \frac{-32 \pm \sqrt{1024 - 960}}{-10} \][/tex]
[tex]\[ x = \frac{-32 \pm \sqrt{64}}{-10} \][/tex]
[tex]\[ x = \frac{-32 \pm 8}{-10} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-24}{-10} = \frac{12}{5} \][/tex]
[tex]\[ x = \frac{-40}{-10} = 4 \][/tex]
So, the solutions to the third equation are:
[tex]\[ x = \frac{12}{5} \quad \text{or} \quad x = 4 \][/tex]
Thus, the solutions are:
1) [tex]\( x = 2 \)[/tex] or [tex]\( x = -\frac{5}{16} \)[/tex]
2) [tex]\( x = \frac{2}{3} \)[/tex] or [tex]\( x = -\frac{1}{3} \)[/tex]
3) [tex]\( x = \frac{12}{5} \)[/tex] or [tex]\( x = 4 \)[/tex]
### 1) [tex]\( 16x - \frac{10}{x} = 27 \)[/tex]
First, multiply both sides of the equation by [tex]\( x \)[/tex] to get rid of the fraction:
[tex]\[ 16x^2 - 10 = 27x \][/tex]
Rearrange the equation to form a standard quadratic equation:
[tex]\[ 16x^2 - 27x - 10 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 16 \)[/tex], [tex]\( b = -27 \)[/tex], and [tex]\( c = -10 \)[/tex]:
[tex]\[ x = \frac{27 \pm \sqrt{(-27)^2 - 4 \cdot 16 \cdot (-10)}}{2 \cdot 16} \][/tex]
[tex]\[ x = \frac{27 \pm \sqrt{729 + 640}}{32} \][/tex]
[tex]\[ x = \frac{27 \pm \sqrt{1369}}{32} \][/tex]
[tex]\[ x = \frac{27 \pm 37}{32} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{64}{32} = 2 \][/tex]
[tex]\[ x = \frac{-10}{32} = -\frac{5}{16} \][/tex]
So, the solutions to the first equation are:
[tex]\[ x = 2 \quad \text{or} \quad x = -\frac{5}{16} \][/tex]
### 2) [tex]\( 9x^2 - 3x - 2 = 0 \)[/tex]
This is already a quadratic equation in standard form. Let's use the quadratic formula again: [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 9 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -2 \)[/tex]:
[tex]\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 9 \cdot (-2)}}{2 \cdot 9} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{9 + 72}}{18} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{81}}{18} \][/tex]
[tex]\[ x = \frac{3 \pm 9}{18} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{12}{18} = \frac{2}{3} \][/tex]
[tex]\[ x = \frac{-6}{18} = -\frac{1}{3} \][/tex]
So, the solutions to the second equation are:
[tex]\[ x = \frac{2}{3} \quad \text{or} \quad x = -\frac{1}{3} \][/tex]
### 3) [tex]\( \frac{1}{x-3} + \frac{2}{x-2} = \frac{8}{x} \)[/tex]
To solve this, first find a common denominator:
[tex]\[ \frac{(x-2) + 2(x-3)}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
[tex]\[ \frac{x - 2 + 2x - 6}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
[tex]\[ \frac{3x - 8}{(x-3)(x-2)} = \frac{8}{x} \][/tex]
Cross-multiply to clear the fractions:
[tex]\[ x(3x - 8) = 8(x-3)(x-2) \][/tex]
[tex]\[ 3x^2 - 8x = 8(x^2 - 5x + 6) \][/tex]
[tex]\[ 3x^2 - 8x = 8x^2 - 40x + 48 \][/tex]
Rearrange the equation:
[tex]\[ 3x^2 - 8x - 8x^2 + 40x - 48 = 0 \][/tex]
[tex]\[ -5x^2 + 32x - 48 = 0 \][/tex]
Solve using the quadratic formula: [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -5 \)[/tex], [tex]\( b = 32 \)[/tex], and [tex]\( c = -48 \)[/tex]:
[tex]\[ x = \frac{-32 \pm \sqrt{32^2 - 4 \cdot (-5) \cdot (-48)}}{2 \cdot (-5)} \][/tex]
[tex]\[ x = \frac{-32 \pm \sqrt{1024 - 960}}{-10} \][/tex]
[tex]\[ x = \frac{-32 \pm \sqrt{64}}{-10} \][/tex]
[tex]\[ x = \frac{-32 \pm 8}{-10} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-24}{-10} = \frac{12}{5} \][/tex]
[tex]\[ x = \frac{-40}{-10} = 4 \][/tex]
So, the solutions to the third equation are:
[tex]\[ x = \frac{12}{5} \quad \text{or} \quad x = 4 \][/tex]
Thus, the solutions are:
1) [tex]\( x = 2 \)[/tex] or [tex]\( x = -\frac{5}{16} \)[/tex]
2) [tex]\( x = \frac{2}{3} \)[/tex] or [tex]\( x = -\frac{1}{3} \)[/tex]
3) [tex]\( x = \frac{12}{5} \)[/tex] or [tex]\( x = 4 \)[/tex]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
