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Sagot :
Certainly! Let's solve the equation step-by-step:
Given the equation:
[tex]$(x + 2)(x + 3) + \sqrt{-3}(x - 2) - 2x(x + 1) = 0$[/tex]
First, we'll expand and simplify each term separately.
1. Expand [tex]\((x + 2)(x + 3)\)[/tex]:
[tex]\[ (x + 2)(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6 \][/tex]
2. Note that [tex]\(\sqrt{-3}\)[/tex] can be written as [tex]\(i\sqrt{3}\)[/tex] (where [tex]\(i\)[/tex] is the imaginary unit), so [tex]\(\sqrt{-3}(x - 2)\)[/tex] becomes:
[tex]\[ i\sqrt{3}(x - 2) = i\sqrt{3}x - 2i\sqrt{3} \][/tex]
3. Expand [tex]\(- 2x(x + 1)\)[/tex]:
[tex]\[ -2x(x + 1) = -2x^2 - 2x \][/tex]
Now we combine all the terms together:
[tex]\[ x^2 + 5x + 6 + i\sqrt{3}x - 2i\sqrt{3} - 2x^2 - 2x = 0 \][/tex]
Combine like terms:
[tex]\[ (x^2 - 2x^2) + (5x - 2x + i\sqrt{3}x) + (6 - 2i\sqrt{3}) = 0 \][/tex]
[tex]\[ -x^2 + (3x + i\sqrt{3}x) + 6 - 2i\sqrt{3} = 0 \][/tex]
[tex]\[ -x^2 + x(3 + i\sqrt{3}) + 6 - 2i\sqrt{3} = 0 \][/tex]
Now, we need to solve this quadratic equation:
[tex]\[ -x^2 + x(3 + i\sqrt{3}) + 6 - 2i\sqrt{3} = 0 \][/tex]
To find the solutions, we use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = -1\)[/tex], [tex]\(b = 3 + i\sqrt{3}\)[/tex], and [tex]\(c = 6 - 2i\sqrt{3}\)[/tex].
Let's solve it step-by-step:
The quadratic formula is:
[tex]\[ x = \frac{-(3 + i\sqrt{3}) \pm \sqrt{(3 + i\sqrt{3})^2 - 4(-1)(6 - 2i\sqrt{3})}}{2(-1)} \][/tex]
First, calculate [tex]\(b^2 = (3 + i\sqrt{3})^2\)[/tex]:
[tex]\[ (3 + i\sqrt{3})^2 = 9 + 6i\sqrt{3} - 3 = 6 + 6i\sqrt{3} \][/tex]
Next, calculate [tex]\(-4ac = -4(-1)(6 - 2i\sqrt{3})\)[/tex]:
[tex]\[ -4(-1)(6 - 2i\sqrt{3}) = 24 - 8i\sqrt{3} \][/tex]
Combine these results:
[tex]\[ 6 + 6i\sqrt{3} + 24 - 8i\sqrt{3} = 30 - 2i\sqrt{3} \][/tex]
Now, the current form of the quadratic formula is:
[tex]\[ x = \frac{-(3 + i\sqrt{3}) \pm \sqrt{30 - 2i\sqrt{3}}}{-2} \][/tex]
So, the quadratic roots are:
[tex]\[ x = \frac{-(3 + i\sqrt{3}) \pm \sqrt{30 - 2i\sqrt{3}}}{-2} \][/tex]
Which simplifies to the solutions:
[tex]\[ x = \frac{3}{2} + \frac{\sqrt{30 - 2\sqrt{3}i}}{2} + \frac{\sqrt{3}i}{2} \][/tex]
and
[tex]\[ x = \frac{3}{2} - \frac{\sqrt{30 - 2\sqrt{3}i}}{2} + \frac{\sqrt{3}i}{2} \][/tex]
Therefore, the solutions to the equation [tex]\((x+2)(x+3)+(\sqrt{-3})(x-2)-2 x(x+1)=0\)[/tex] are:
[tex]\[ x = \frac{3}{2} + \frac{\sqrt{30 - 2i\sqrt{3}}}{2} + \frac{i\sqrt{3}}{2} \][/tex]
and
[tex]\[ x = \frac{3}{2} - \frac{\sqrt{30 - 2i\sqrt{3}}}{2} + \frac{i\sqrt{3}}{2} \][/tex]
Given the equation:
[tex]$(x + 2)(x + 3) + \sqrt{-3}(x - 2) - 2x(x + 1) = 0$[/tex]
First, we'll expand and simplify each term separately.
1. Expand [tex]\((x + 2)(x + 3)\)[/tex]:
[tex]\[ (x + 2)(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6 \][/tex]
2. Note that [tex]\(\sqrt{-3}\)[/tex] can be written as [tex]\(i\sqrt{3}\)[/tex] (where [tex]\(i\)[/tex] is the imaginary unit), so [tex]\(\sqrt{-3}(x - 2)\)[/tex] becomes:
[tex]\[ i\sqrt{3}(x - 2) = i\sqrt{3}x - 2i\sqrt{3} \][/tex]
3. Expand [tex]\(- 2x(x + 1)\)[/tex]:
[tex]\[ -2x(x + 1) = -2x^2 - 2x \][/tex]
Now we combine all the terms together:
[tex]\[ x^2 + 5x + 6 + i\sqrt{3}x - 2i\sqrt{3} - 2x^2 - 2x = 0 \][/tex]
Combine like terms:
[tex]\[ (x^2 - 2x^2) + (5x - 2x + i\sqrt{3}x) + (6 - 2i\sqrt{3}) = 0 \][/tex]
[tex]\[ -x^2 + (3x + i\sqrt{3}x) + 6 - 2i\sqrt{3} = 0 \][/tex]
[tex]\[ -x^2 + x(3 + i\sqrt{3}) + 6 - 2i\sqrt{3} = 0 \][/tex]
Now, we need to solve this quadratic equation:
[tex]\[ -x^2 + x(3 + i\sqrt{3}) + 6 - 2i\sqrt{3} = 0 \][/tex]
To find the solutions, we use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = -1\)[/tex], [tex]\(b = 3 + i\sqrt{3}\)[/tex], and [tex]\(c = 6 - 2i\sqrt{3}\)[/tex].
Let's solve it step-by-step:
The quadratic formula is:
[tex]\[ x = \frac{-(3 + i\sqrt{3}) \pm \sqrt{(3 + i\sqrt{3})^2 - 4(-1)(6 - 2i\sqrt{3})}}{2(-1)} \][/tex]
First, calculate [tex]\(b^2 = (3 + i\sqrt{3})^2\)[/tex]:
[tex]\[ (3 + i\sqrt{3})^2 = 9 + 6i\sqrt{3} - 3 = 6 + 6i\sqrt{3} \][/tex]
Next, calculate [tex]\(-4ac = -4(-1)(6 - 2i\sqrt{3})\)[/tex]:
[tex]\[ -4(-1)(6 - 2i\sqrt{3}) = 24 - 8i\sqrt{3} \][/tex]
Combine these results:
[tex]\[ 6 + 6i\sqrt{3} + 24 - 8i\sqrt{3} = 30 - 2i\sqrt{3} \][/tex]
Now, the current form of the quadratic formula is:
[tex]\[ x = \frac{-(3 + i\sqrt{3}) \pm \sqrt{30 - 2i\sqrt{3}}}{-2} \][/tex]
So, the quadratic roots are:
[tex]\[ x = \frac{-(3 + i\sqrt{3}) \pm \sqrt{30 - 2i\sqrt{3}}}{-2} \][/tex]
Which simplifies to the solutions:
[tex]\[ x = \frac{3}{2} + \frac{\sqrt{30 - 2\sqrt{3}i}}{2} + \frac{\sqrt{3}i}{2} \][/tex]
and
[tex]\[ x = \frac{3}{2} - \frac{\sqrt{30 - 2\sqrt{3}i}}{2} + \frac{\sqrt{3}i}{2} \][/tex]
Therefore, the solutions to the equation [tex]\((x+2)(x+3)+(\sqrt{-3})(x-2)-2 x(x+1)=0\)[/tex] are:
[tex]\[ x = \frac{3}{2} + \frac{\sqrt{30 - 2i\sqrt{3}}}{2} + \frac{i\sqrt{3}}{2} \][/tex]
and
[tex]\[ x = \frac{3}{2} - \frac{\sqrt{30 - 2i\sqrt{3}}}{2} + \frac{i\sqrt{3}}{2} \][/tex]
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