A general quadratic equation is of the form ax^{2} + bx + c = 0 where "a" is not equal to 0. The easy method of getting the roots of the equation is by using quadratic formula.

A **complex number** is a number consisting of a real and imaginary part. It can be written in the form a + bi, where a and b are real numbers, and i is the standard imaginary unit with the property i ^{2} = -1.

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A polynomial P(x) of degree n has exactly n roots, real or complex. Any polynomial with complex coefficients it has exactly n complex roots. Roots of quadratic polynomial are imaginary if the value of discriminant is less than zero.

A polynomial of degree n has at least one root, real or complex.

A quadratic formula is used to find the roots of any quadratic equation. If quadratic equation in the form ax^{2} + bx + c = 0, then quadratic formula for x has the form:

x = $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $b^{2}$ - 4ac is discriminant denoted by $\Delta$.

Nature of roots depends upon discriminant:

If $\Delta$ > 0, then the roots are real and unequal.

If $\Delta$ = 0, then the roots are real and equal.

If $\Delta$ < 0, then the roots are imaginary.

Imaginary roots appear in a quadratic equation when the discriminant ($\Delta$) of the quadratic equation is -ve.

Let us find the roots of the following equation x^{2} + 256 = 0

**Step 1:** Write the given equation .

The given equation is, x2 + 256 = 0

**Step 2:** Subtract 256 on both sides.

x^{2} = -256

**Step 3:** Take square root on both sides.

x = $\pm \sqrt{-256}$

x = ± 16 $\sqrt{-1}$

x = + 16i, - 16i

Therefore, given equation contain two complex roots, x = 16i and x = -16i. Given below are some of the examples on finding imaginary roots.### Solved Examples

**Question 1: **Find the roots of the following equation x^{2} + 9 = 0

** Solution: **

**Question 2: **Find the roots of the following equation x^{2} + 25 = 0

** Solution: **

A polynomial of degree n has at least one root, real or complex.

A quadratic formula is used to find the roots of any quadratic equation. If quadratic equation in the form ax

x = $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $b^{2}$ - 4ac is discriminant denoted by $\Delta$.

Nature of roots depends upon discriminant:

If $\Delta$ > 0, then the roots are real and unequal.

If $\Delta$ = 0, then the roots are real and equal.

If $\Delta$ < 0, then the roots are imaginary.

Imaginary roots appear in a quadratic equation when the discriminant ($\Delta$) of the quadratic equation is -ve.

The given equation is, x2 + 256 = 0

x

x = $\pm \sqrt{-256}$

x = ± 16 $\sqrt{-1}$

x = + 16i, - 16i

Therefore, given equation contain two complex roots, x = 16i and x = -16i. Given below are some of the examples on finding imaginary roots.

** Step 1:** Write the given equation

The given equation is, x^{2} + 9 = 0** Step 2: **Subtract 9 on both sides.

x^{2} = -9

** Step 3:** Take square root on both sides.

$x = \pm \sqrt{-9}$

$x = \pm 3 \sqrt{-1}$

x = + 3i, -3i

** Step 4:** Write the solution.

Therefore, the given equation contains two complex roots, x = 3i and x = -3i.

** Step 1:** Write the given equation

The given equation is x2 + 25 = 0**Step 2:** Subtract 25 on both sides.

x^{2} = -25**Step 3:** Take square root on both sides.

$x = \pm \sqrt{-25}$

$x = \pm 5 \sqrt{-1}$

x = + 5i, -5i

** Step 4: **Write the solution.

Therefore, given equation contain two complex roots, x = 5i and x = -5i.

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