`star` Quadratic Equations

● We are already familiar with the quadratic equations and have solved them in the set of real numbers in the cases where `color(blue)("discriminant is non-negative")`, i.e., ≥ 0,

● Let us consider the following quadratic equation: `color(blue)(ax^2 + bx + c = 0)` with real coefficients `a, b, c` and `a ≠ 0.`

Also, let us assume that the `color(red)(b^2 – 4ac < 0.)`

● Now, we know that we can find the square root of negative real numbers in the set of complex numbers.

Therefore, the solutions to the above equation are available in the set of complex numbers which are given by

`color(red)(x =(-bpm sqrt(b^2-4ac))/(2a)=(-bpm sqrt(4ac-b^2) * i)/(2a))`

Note :

`color(blue)("“A polynomial equation has at least one root.”")`

`color(blue)("“A polynomial equation of degree n has n roots.” ")`

● Let us consider the following quadratic equation: `color(blue)(ax^2 + bx + c = 0)` with real coefficients `a, b, c` and `a ≠ 0.`

Also, let us assume that the `color(red)(b^2 – 4ac < 0.)`

● Now, we know that we can find the square root of negative real numbers in the set of complex numbers.

Therefore, the solutions to the above equation are available in the set of complex numbers which are given by

`color(red)(x =(-bpm sqrt(b^2-4ac))/(2a)=(-bpm sqrt(4ac-b^2) * i)/(2a))`

`color{red} "Key Point"` - At this point of time, some would be interested to know as to how many roots does an equation have? In this regard, the following theorem known as the Fundamental theorem of Algebra is stated below.

Note :

`color(blue)("“A polynomial equation has at least one root.”")`

`color(blue)("“A polynomial equation of degree n has n roots.” ")`

Q 3018012809

Solve `x^2 + 2 = 0`

We have, `x^2 + 2 = 0`

or `x^2 = -2`

`x = pm sqrt2 = pm sqrt2i`

Q 3038112902

Solve `x^2 + x + 1= 0`

Here, `b^2 – 4ac = 12 – 4 × 1 × 1 = 1 – 4 = – 3`

Therefore, the solutions are given by `x = (-1 pm sqrt(-3))/(2xx1) = (-1 pm sqrt3 i)/2`

Q 3068112905

Solve `sqrt5 x^2 +x+ sqrt5 = 0`

Here, the discriminant of the equation is

`1^2-4xx sqrt5 xx sqrt5 = 1-20 = -19`

Therefore, the solutions are `(-1 pm sqrt(-19))/(2 sqrt5) = (-1 pm sqrt(19i))/(2 sqrt5)`