Nature of roots in Quadratic equations

For a.x^2 + b.x + c = 0 to have rational roots, necessary and sufficient conditions are:
1. b/a and c/a must be rational.
2. Discriminant must be square of a rational number.


Similarly for integer roots:
1. b/a and c/a must be integers.
2. Discriminant must be perfect square of an integer.

Let's go in some more detail here.
If b/a and c/a are integers I can rewrite the equation as:
x^2 + B.x + C = 0
where B = b/a and C = c/a.
roots = {-B +- sqrt(B^2 - 4C)}/2
Case 1: 
B is even.
roots = (even +- even)/2 = integer

Case 2:
B is odd.
roots = (odd +- odd)/2 = integer

So yeah, we are good.

Ex1:

Find possible values of $k$ such that
$x^2 - kx + 4 = 0$
has integer roots, given that $k \in \{-10, -9, -8, \ldots, 9, 10\}$

Solution 1:
Discriminant >= 0 and Perfect square of an integer
=> k^2 - 16 = m^2 where m is an integer.
(k+m).(k-m) = 16
Now LHS can take on factors of 16:
1,16
2,8
4,4
8,2
16,1
-1,-16
-2,-8
-4,-4
-8,-2
-16,-1
Possible Integer values of (k,m) are:
5,3 | 5,-3
4,0
-5,3 | -5,-3
-4,0

So unique values of k: 5,-5,4,-4

Solution 2:
If the 2 roots are a,b then 
a + b = -k
a.b = 4
a,b are integers => 
(a,b) = (1,4)(2,2)(4,1)(-1,-4)(-2,-2)(-4,-4)
=> k = 1+4,2+2,-4,-5


Example 2: 

$a, b \in \mathbb{Z}$ and $a + b$ is root of
$x^2 + ax + b = 0$
Find max. possible value of $b^2$?

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Solution:

Given:

$$a + b \rightarrow x^2 + ax + b = 0$$

Substitute $a + b$ into the equation:

$$(a + b)^2 + a(a + b) + b = 0$$

Expand:

$$a^2 + b^2 + 2ab + a^2 + ab + b = 0$$

Combine like terms:

$$2a^2 + 3ab + b^2 + b = 0$$

Rewriting:

$$2a^2 + (3b)a + b^2 + b = 0$$

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Now we have got a quadratic equation in 'a'.
So
1) -3b/2 and (b^2 + b)/2 should be integers.
2) Discriminant should be square of an integer.

$$\frac{9b^2}{4} - \frac{4(b^2 + b)}{2} = m^2$$

Simplifies to:

$$9b^2 - 8b^2 - 8b = 4m^2$$

Or more cleanly:

$$b^2 - 8b = 4m^2$$

$$b^2 - 8b + 16 = 4m^2 + 16$$ $$(b - 4)^2 = 4m^2 + 16$$ $$(b - 4)^2 - 4m^2 = 16$$ $$(b - 4 - 2m)(b - 4 + 2m) = 16$$

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Again we factorize like earlier.
And max value of b^2 will be 64.
Remember to satisfy 1) above also.

Here is the transcribed text from the image:

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Ex3:
$a, b, c$ are real & distinct numbers,
such that $a + b + c = 0$, then
the roots of equation

$$3a x^2 + 5b x + 7c = 0$$

are ___

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a) Real & distinct
b) Imaginary
c) Equal
d) Negative

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Solution:
Discriminant of the given quadratic can be written in 2 ways:

$$D = 25(a + c)^2 - 84ac$$ $$= 25a^2 + 25c^2 + 50ac - 84ac$$ $$= 25a^2 + 25c^2 - 34ac$$ $$= 25a^2 + 25c^2 - 50ac + 16ac$$ $$D = 25(a - c)^2 + 16ac$$

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Use the first form of D to deduce that D > 0 when ac < 0.
Use the second form of D to deduce that D > 0 when ac > 0.
=> D > 0 always.
So roots are real and distinct.

Ex4:

Find solution of

$$(x - 1)^2 - 3|x - 1| - 28 = 0$$

Definition of absolute value:

$$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$$

Let $|x - 1| = t$, then:

$$t^2 - 3t - 28 = 0$$

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Solving this quadratic, we get:
t = 7,-4
But t = |x-1| means it can't be negative.
=> t = 7 => x = -6,8

Range of quadratic function:
y = ax^2 + bx + c
a > 0 => upward parabola.
=> no max value, only min value.
Min value will occur at vertex.
Vertex is in between the 2 roots.
1/2{-2b/2a} = -b/2a
Once you put this value, you will get:
y_min = -D/4a

Similarly when a < 0
y_max = -D/4a

Ex5:

Find max value of

$$y = \frac{1}{x^2 - 2x + 3}$$

Analyzing the denominator:

$$x^2 - 2x + 3$$ $$\text{min} = -\frac{(4 - 12)}{4} = 2, \quad \text{max} = \infty$$ $$y_{\text{max}} = \frac{1}{\min(x^2 - 2x + 3)}$$ $$y_{\text{max}} = \frac{1}{2}$$

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Ex6:

$$x^2 - 3x + 5 = 0 \quad \text{roots are } \alpha \text{ and } \beta$$

Find sum of roots of the quadratic whose roots are

$$\alpha^3 + 3\alpha - 7 \quad \text{and} \quad \beta^4 - 4\beta^3 + 7\beta + 3$$

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Solution:
High powers of α and β can be made linear by plugging them into the original equation.
Eventual answer should be 42.

Ex7:

If $\alpha$ & $\beta$ are roots of equation

$$(x - \alpha)(x - \beta) = c$$

then roots of quadratic

$$(x - \alpha)(x - \beta) + c = 0$$

are ___

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Solution:
Using the given equation:
(x-a).(x-b) - c =(x−α)(x−β)
=> (x−α)(x−β)+c= (x-a).(x-b)
Answer: a,b

Ex8:

If $\alpha$ & $\beta$ are roots of

$$x^2 - 7x + 9 = 0$$

and

$$P_n = \alpha^n + \beta^n$$

Find the value of

$$\frac{9P_{10} + P_{12}}{28P_{11}}$$

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Solution:

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Given:

$$\alpha^2 - 7\alpha + 9 = 0$$

Multiply both sides by $\alpha^{n-2}$:

$$\alpha^{n-2}(\alpha^2 - 7\alpha + 9) = \alpha^{n-2} \cdot 0$$

Which leads to:

$$\alpha^n - 7\alpha^{n-1} + 9\alpha^{n-2} = 0$$

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From earlier steps:

$$\alpha^n - 7\alpha^{n-1} + 9\alpha^{n-2} = 0$$ $$\beta^n - 7\beta^{n-1} + 9\beta^{n-2} = 0$$

Add the two equations:

$$\alpha^n + \beta^n - 7(\alpha^{n-1} + \beta^{n-1}) + 9(\alpha^{n-2} + \beta^{n-2}) = 0$$

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This simplifies to the recurrence relation for $P_n$:

$$P_n = 7P_{n-1} - 9P_{n-2}$$

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Using the recurrence relation:

$$P_n - 7P_{n-1} + 9P_{n-2} = 0$$

Apply it for $n = 12$:

$$P_{12} - 7P_{11} + 9P_{10} = 0$$

Rearranged:

$$P_{12} + 9P_{10} = 7P_{11}$$

Divide both sides by $28P_{11}$:

$$\frac{P_{12} + 9P_{10}}{28P_{11}} = \frac{7}{28} = \frac{1}{4}$$

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So, the final answer is:

$$\boxed{{\displaystyle \frac{1}{4}}}$$