Find the range of 'a' in the following equations by the number of real roots.

Question

Greetings! Can anyone guide me on how to solve this type of questions? My teacher already gave the solutions for the questions but I can't understand them at all :/

Here's the link to the questions and answers given by my teacher:

The questions - https://i.imgur.com/dzF7cgU.jpeg

Answer for number 1 and 2 - https://i.imgur.com/4ZSaKFx.jpeg

Answer for number 3 - https://i.imgur.com/QpW1OTY.jpeg

Answer for number 4 and 5 - https://i.imgur.com/jqNKHkH.jpeg

Thank you so much!

Answer 1

The teacher is asking

for which values of 'a' the equation will have one real root,

for which values of 'a' it will have two real roots, and

for which values of 'a' it will have no real root.

His answers involve different ways of identifying the roots, but

here is a general approach.

The general formula for the roots of a quadratic equation involves a square root.

If the quantity under the square root is negative then there is no real root.

If the quantity under the square root is positive then there are two real roots.

If the quantity under the square root is 0 then there is one real root.

In your four examples, the quantities under the square root are

1. 16a² - 16a²

2. 4a² - 12a + 8

3. 16 - 16a

4. 9 - 8a

In problem 1. you are taking square root of 0, therefore

for all values of 'a' the equation has 1 root.

In problem 2. you need to ask for which 'a'

4a² - 12a + 8 > 0 (or equal 0)

The parabola 4a² - 12a + 8 has value 0 for a = 1 and for a = 2.

Therefore the equation of problem 2. will be only a single real root when a = 1 or a = 2.

The parabola 4a² - 12a + 8 is negative for 1 < a < 2;

therefore for these values of 'a' the given equation has no real root.

For all other values of 'a' the equation has two real roots.

For problems 3. and 4. you simply need to solve the inequalities

16 - 16a > 0 and 9 - 8a >0, respectively.