$$a+b = t-3$$ and $$ab = -2t+1$$
So,$$a^2+b^2 = (t-3)^2 +4t-2 = t^2-2t+7 =(t-1)^2 + 6$$
So, the minimum value of $$a^2+b^2=6$$ when t=1

Let the roots of the equation $$x^2-ax+b=0$$ be $$x_1$$ and $$x_2$$.
So, $$\frac{x_1+x_2}{x_1 * x_2} = 2$$
So, $$\frac{a}{b}=2$$
Similarly, the sum of the reciprocals of the roots of the quadratic equation $$x^2-bx+a=0$$ is $$\frac{b}{a}$$ and equals 0.5

Consider the expression $$x(x+1)(x+2)(x+3) + 1$$

$$x*(x+1)*(x+2)*(x+3) + 1 = (x^2+3x)*(x^2+3x+2)+1$$

This is of the form $$a(a+2) + 1$$ where $$a = x^2 + 3x$$
But, $$a(a+2)+1 = a^2 + 2a + 1 = (a+1)^2$$

Therefore, the given expression is equal to $$(x^2+3x+1)^2$$

Substituting $$x$$ = $$2014$$ in the equation $$(x^2+3x+1)^2$$ , we get,

So, $$N = 2014^2+3*2014+1=4062239$$

From the sum and product of the roots of the equation, we know that

p+q=-b/a and pq=c/a.
Hence the new equation will have roots with product = 1 and
sum = $$p/q+q/p = (p^2+q^2)/pq = ((p+q)^2-2pq)/pq = (-b/a)^2-2c/a)/c/a = (b^2-2ac)/ac$$ .

Hence the equation is $$x^2-((b^2-2ac)/ac) x+1=0$$

Let the roots be 'a' and 'b'
So, $$a+b=1$$ and $$ab=-P$$ and $$a^4+b^4=337$$
$$a^2+b^2 = 1+2P$$.
$$a^4+b^4 = (2P+1)^2-2P^2 = 2P^2+4P+1$$
So, $$P^2+2P-168=0$$ or $$P= 12$$ or $$-14$$
If $$P=-14$$, the roots of the equation are not real
So, $$P=12$$

Let the quadratic equation f(x) be $$ax^2+bx+c = 0$$
So to find the value of x at which attains minimum value, differentiating f(x) we get 2ax+b=0  => x=$$\frac{-b}{2a}$$

Here =$$\frac{-b}{2a}$$ = -2 

Now the sum of the roots = $$\frac{-b}{a}$$ =2*$$\frac{-b}{2a}$$ = 2*(-2)=-4

From the properties of quadratic equations,
x+y = 5 and xy = -1.
So, (4+x)*(4+y) = 16 + 4*(x+y) + xy = 16+ 4*5 -1 = 35

The product of the roots $$= -30/15=-2$$.
Hence the third root $$= -2$$.
Let the other two roots be p and 1/p.
Hence $$p+1/p-2=4/15$$ => $$p+1/p=34/15$$ => $$15p^2-34p +15=0$$.
The two roots are 3/5 and 5/3

For the roots of a quadratic equation to be real the discriminant should be greater than or equal to zero.
So, $$13^2-4\times2\times|a|\geq0$$
=> $$|a|\leq\frac{169}{8}$$
So ‘a’ can take values from -21, -20, … to 21.
Thus the number of integral values ‘a’ can take is 43.

From the first equation, the sum of the roots is 15 and from the second equation, the product of the roots is 50.

Therefore the quadratic equation is $$x^2-15x+50 = 0$$.

The roots are 5,10.

$$x^2-(m-3)x+(m-8)=0$$
Let the roots be a and b.
$$a^2+b^2 = (a+b)^2-2ab = (m-3)^2-2(m-8)$$ = $$m^2-8m+25$$
In order to find the minimum value of a quadratic equation $$ax^2+bx+c$$, we need to convert it into a square function, we need to convert it into the form of $$(x-d)^2+e$$
This is because we know that the minimum value of a square function is always 0.
Hence, $$m^2-8m+25$$ = $$(m-4)^2+9$$
Minimum value would be 9.

The roots of the equation $$x^2+9x+|k|=0$$ are $$\frac{-9\pm\sqrt{81-4|k|}}{2}$$. For these roots to be integers $$81-4|k|$$ should be the square of an odd integer.
$$81-4|k|$$ can take the values of 49, 25, 9, 1.
Also when k=0, the roots are integers.
So the possible values of ‘k’ are $$0,\pm8,\pm14,\pm18,\pm20$$.
Thus there are 9 possible values in total.

We know that p+q = -b/a and pq=c/a.
Therefore: $$-p^2q^2= -c^2/a^2$$ and $$p^2-q^2=(p+q)(p-q)=(-b/a)*1$$.
Hence the eqn is :$$ x^2-(-b/a)x-c^2/a^2=0$$ => $$a^2x^2 + abx - c^2=0$$

Given expression = (x-5)(x-10)(x-15)..... and so on till (x-60)
The highest power in the expression is $$x^{12}$$ and its coefficient is 1.
Therefore $$x^{11}$$ will give the negative value of sum of all the possible roots of the equation.
Coefficient of $$x^{11}$$ = -(5+10+15+....60) = -(6*65) = -390.

-q/r = 10 and -p/q = 20. The product of the roots of the equation $$rx^2+qx+p=0$$ is p/r which equals 20*10=200

The equation of the form $$y = x^2 + ax + b$$ is an upward facing parabola anywhere on the 2-D plane.

The quadratic equation $$x^2 + ax + b$$ is positive for all values of x except when x=3 only if the quadratic equation is $$y = (x-3)^2$$.

Hence, $$ y = = (x-3)^2 = x^2-6x+9\ \longrightarrow\ a = -6$$ and $$b = 9$$

So, $$x^2 + 2ax + 4b$$ can be represented as $$y = x^2-12x+36=(x-6)^2$$

$$y = (x-6)^2$$ is 0 at $$x = 6$$, however, it is positive every where else.

Therefore $$y = (x-6)^2$$ has 0 negative values.

g(x) has a maximum value. So it is a downward open parabola. But nothing can be said about f(x) except that it has the same roots as g(x). It can have a positive leading coefficient, in which case the function takes a minimum value or it can have a negative leading coefficient, in which case the function takes a maximum value. But nothing can be said about the magnitude of the value.

An example of upward and downward opening parabola with same roots is shown in the following graph:

The upward opening parabola equation : $$y\ =6\left(x^2-3x+2\right)$$

The downward opening parabola equation : $$y\ =-x^2+3x-2$$

Let the roots of the first equation be x, 3x
Let the roots of the second equation be 3y, 5y
It is given that,
x+3x = 3y+5y
=> 4x = 8y
=> x = 2y

So the roots of the first equation will be 2y, 6y

The difference in products of both the equations is,
$$(3y)(5y) - (2y)(6y) = 15y^2 - 12y^2=3y^2$$

If y=1, then the difference will be lowest which is 3.

For the equation to have real roots, $$(k-4)^2 >= 4(k+4) => k <= 0$$ and $$k >= 12$$.

Sum of the roots = -b/a = 4-k. Product of the roots = k+4.
Sum of squares of the roots = $$(4-k)^2 -2*(k+4) = k^2 -10k + 8.$$

This obtains the least value at -b/2a => at k = 5.
But since k cannot take the value 5, the least value of the expression is obtained at k=0, and the value is 8.
 

From the first equation, the sum of the roots is 15 and from the second equation, the product of the roots is 50.

Therefore the quadratic equation is $$x^2-15x+50 = 0$$.

The roots are 5,10.

As the function intersects Y axis at 2, d = 2.
As the function intersects X axis at 1 and -1,
a+b+c+d = 0 and -a + b -c + d =0
So, b+d =0 or b = -2