(Q26-Q30) : (1 - 2) Introduction in Complex Numbers.

 

 

1sec ≫ First term  ≫ Algebra ≫ Unit 1 ≫ (1-2 )  introduction to complex numbers :

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(Q20 To Q25)

[26] Solve in the set of complex numbers:

$3x^{2} + 12 = 0$

First, we divide the equation by 3 to get a simpler equation:

\[ x^{2} + 4 = 0 \]

\[ x^{2} = -4 \]

\[ x = \pm \sqrt{-4} = \pm 2i \]

So, the solution in the set of complex numbers is:

\[ x = 2i \quad \text{or} \quad x = -2i \]

[27] Solve in the set of complex numbers:

$4x^{2} + 100 = 75$

\[ 4x^{2} = -25 \]

\[ x^{2} = -\frac{25}{4} \]

\[ x = \pm \sqrt{-\frac{25}{4}} = \pm \frac{5i}{2} \]

\[ x = \frac{5i}{2} \quad \text{or} \quad x = -\frac{5i}{2} \]

[28] Solve in the set of complex numbers:

$x^{2} - 4x + 5 = 0$

First, use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 1\), \(b = -4\), and \(c = 5\). Substitute these values into the formula:

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \]

Simplify the expression inside the square root:

\[ x = \frac{4 \pm \sqrt{16 - 20}}{2} \]

\[ x = \frac{4 \pm \sqrt{-4}}{2} \]

Since \(\sqrt{-4} = 2i\), we have:

\[ x = \frac{4 \pm 2i}{2} \]

Simplify the fraction:

\[ x = 2 \pm i \]

So, the solution in the set of complex numbers is:

\[ x = 2 + i \quad \text{or} \quad x = 2 - i \]

[29] Solve in the set of complex numbers:

$2x^{2} + 6x + 5 = 0$

To solve the equation \(2x^{2} + 6x + 5 = 0\) in the set of complex numbers, follow these steps:

First, use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 2\), \(b = 6\), and \(c = 5\). Substitute these values into the formula:

\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 2 \cdot 5}}{2 \cdot 2} \]

Simplify the expression inside the square root:

\[ x = \frac{-6 \pm \sqrt{36 - 40}}{4} \]

\[ x = \frac{-6 \pm \sqrt{-4}}{4} \]

Since \(\sqrt{-4} = 2i\), we have:

\[ x = \frac{-6 \pm 2i}{4} \]

Simplify the fraction:

\[ x = \frac{-6}{4} \pm \frac{2i}{4} = -\frac{3}{2} \pm \frac{i}{2} \]

So, the solution in the set of complex numbers is:

\[ x = -\frac{3}{2} + \frac{i}{2} \quad \text{or} \quad x = -\frac{3}{2} - \frac{i}{2} \]

[30] Find the values of \(x\) and \(y\) that satisfy each of the following equation where \(x\) and \(y\) are real numbers:

\((2x - y) + (x - 2y)i = 5 + i\)

To find the values of \(x\) and \(y\) that satisfy the equation \((2x - y) + (x - 2y)i = 5 + i\), follow these steps:

First, equate the real and imaginary parts on both sides of the equation:

Real part: \(2x - y = 5\)

Imaginary part: \(x - 2y = 1\)

Now we have a system of linear equations:

\[ \begin{cases} 2x - y = 5 \\ x - 2y = 1 \end{cases} \]

\[ x = 3, \quad y = 1 \]