When Does A Quadratic Have No Solut...
When Does A Quadratic Have No Solution?

Quadratic equations with complex coefficients

The quadratic equations with complex coefficients that means the coefficients of the equations are not real numbers, they may be an imaginary numbers(i).
Quadratic equation as,
$ax^{2} + bx + c = 0$ where a,b,c are complex numbers and a$\ne$ 0.
Roots of this quadratic equation is same as normal quadratic equation. They are given by
$\alpha = \frac{-b + \sqrt{b^{2}-4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

(i) if D = $b^{2}$ - 4ac = 0 then the roots are complex and equal.
(ii) if D = $b^{2} - 4ac \ne$ 0 then complex but unequal roots.
Note : Complex roots of an equation with real coefficients always occur in pairs. However, this may not be true in case of equations with complex coefficients.
For example: $x^{2}$ - 2ix - 1 = 0 has both roots equal to 'i'.

Examples on quadratic equations with complex coefficients

1) Solve the following quadratic equations by factorization method.
(i) $x^{2}$ - 5ix - 6 = 0
Solution : Here a = 1, b = -5i (complex number) and c = -6
$\alpha = \frac{-b + \sqrt{b^{2}-4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

$\alpha = \frac{-(-5i) + \sqrt{(-5i)^{2}-4\times1\times(-6)}}{2\times 1}$

= $\frac{5i + \sqrt{25i^{2}+24}}{2}$

= $\frac{5i + i}{2}$
= 3i
$\beta = \frac{-(-5i) - \sqrt{(-5i)^{2}-4\times1\times(-6)}}{2\times 1}$

= $\frac{5i - \sqrt{25i^{2}+24}}{2}$

= $\frac{5i - i}{2}$
= 2i
So, the roots of the given equation are 3i and 2i.

(ii) $x^{2}$ + 10ix - 21 = 0
Solution : Here a = 1, b = 10i (complex number) and c = -21
$\alpha = \frac{-b + \sqrt{b^{2}-4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

$\alpha = \frac{-10i + \sqrt{(10i)^{2}-4\times1\times(-21)}}{2\times 1}$

= $\frac{-10i + \sqrt{100i^{2}+84}}{2}$

=$\frac{-10i + \sqrt{-16}}{2}$

= $\frac{-10i + 4i}{2}$
= -3i
$\beta = \frac{-10i - \sqrt{(10i)^{2}-4\times1\times(-21)}}{2\times 1}$

= $\frac{-10i - \sqrt{100i^{2}+84}}{2}$

=$\frac{-10i - \sqrt{-16}}{2}$

= $\frac{-10i - 4i}{2}$
= -7i
So, the roots of the given equation are -3i and -7i.

(iii) i$x^{2}$ - x + 12i = 0
Solution : Here a = i (complex number), b = -1 and c = 12i (complex number)
$\alpha = \frac{-b + \sqrt{b^{2}-4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

$\alpha = \frac{-(-1) + \sqrt{1^{2}-4\times i\times(12i)}}{2\times i}$

= $\frac{1 + \sqrt{1-48i^{2}}}{2i}$

=$\frac{1 + \sqrt{1+48}}{2i}$

= $\frac{1 + 7}{2i}$
= $\frac{4}{i}$
= -4i (after rationalizing)
$\beta = \frac{-(-1) - \sqrt{1^{2}-4\times i\times(12i)}}{2\times i}$

= $\frac{1 - \sqrt{1-48i^{2}}}{2i}$

=$\frac{1 - \sqrt{1+48}}{2i}$

= $\frac{1 - 7}{2i}$
= $\frac{-3}{i}$
= 3i (after rationalizing)
So, the roots of the given equation are -4i and 3i.



11th grade math

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