Question

In a quadratic equation a²+bx+c=0, what types of roots are there when
1) D>0
2)D<0​

Answer 1

Answer:

When:-

1] D>0

Equation have two distinct roots.

2] D<0

Equation have no real roots.

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Answer 2

Solution :

For a quadratic equation , ax² + bx + c = 0 , let us derive the roots and the value of the discriminant first. It's important to prove the nature of the roots.

Given quadratic equation -

ax² + bx + c = 0

Lets try completing the square first . The coefficient of x² has to be 1. To make this possible , let's divide both sides by a.

This becomes -

> x² + (b/a) x + (c/a) = 0. [ Step 1]

This has started to take formation and from the above step , you can derive the formulas for the sum and products of roots .

Anyway moving on . .

> x² + (b/a) x = ( -c/a) [ Shifted the c/a term to the RHS as it is a constant ] [ Step 2]

Now , we have to find the value of the LHS constant , which will manage the central 2ab term .

Here , the central term coefficient is b/a . This can be written as 2 (1) ( b/2a) . x

So , as the coefficient of x² is 1, we gave got the LHS constant. It's b²/4a² .

Now to bring the into the equation, add b²/4a to both sides .

> x² + (b/a) x + (b²/4a²) = (-c/a) + (b²/4a²)

> [ x + b/2a ]² = [ b²/4a² - c/a ] [ Step 3]

Now in the RHS , the denominators are a² and a. Taking LCM

RHS -

> [ b² - 4ac ]/[ 4a²] [ Step 4]

The term on the LHS is a perfect square . So, taking a square root both sides ;

> ( x + b/2a) = ± ( √[b²-4ac]/2a) [ Step 5]

,,

Moving the b/2a term to the RHS -

> x = ± ( √[b²-4ac]/2a) - b/2a

> x = [ -b ± √(b²-4ac) ]/2a. [ Step 6]

What we derived till now is a solution for a generalised quadratic equation . Here , the numerator of the term in the third step is a constant , which is called the discriminant or in short D .

We got that D = √b² - 4ac

The value of D can be > , = and < 0 [ That is why the ± sign Is given. ]

Let us move on to the nature of the roots for the following three values of D.

1. D > 0

2. D = 0

3. D < 0

First Case -

For the first case , the Discriminant is greater than 0.

b² - 4ac > 0

: b² > 4ac

: b > 2√ac [ This term will be only positive ]

: b/2 > √ac

From this , we can get brief details about the nature of the roots .

There will be two distinct and real roots . The graph will intersect the x axis at two different and district points . It will be an upwards parabola .

The graphs for all the three cases are given in the attachment below . Don't forget to check them out.

Second case -

b² - 4ac = 0

: b² = 4ac

: b = 2√ac .

This is a special case and here it's clear that the roots are overlapping each other. The equation will have two real and equal roots . The graph will intersect the x axis at only one point.

Third case -

b² - 4ac < 0

: b² < 4ac

: b < 2√ac

There are no real roots . The graph will be lying in the negative X axis and won't intersect the x axis .

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Note - Wether the graphs will be upward or downward parabolas depends on the value of a too .

There will be four cases for that and it is explained in the attachment .

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