CAT Quadratic Equations Questions

We have two equations,

$$4(x^{2}+y^{2}+z^{2}) = a$$ ---(1)

$$4(x - y - z) = 3 + a$$ ---(2)

Substituting the value of a from equation 1 in equation 2, we get,

$$4\left(x\ -\ y\ -\ z\right)\ =\ 3\ +\ 4(x^2\ +\ y^2\ +\ z^2)$$

$$\ 3\ +\ 4(x^2\ +\ y^2\ +\ z^2)\ -4\left(x\ -\ y\ -\ z\right)\ =\ 0$$

$$\ 3\ +\ 4x^2\ +\ 4y^2\ +\ 4z^2\ -4x\ \ +\ 4y\ +\ 4z\ =\ 0$$

It can be written as,

$$4x^2\ -4x\ +\ 1+\ 4y^2\ +\ 4y\ +\ 1\ +\ 4z^2\ +\ 4z\ +\ 1\ =\ 0$$

$$\left(2x\ -\ 1\right)^2\ +\ \left(2y\ +\ 1\right)^2\ +\ \left(2z\ +\ 1\right)^2\ \ =0$$

We know that if the sum of the squares of terms is 0, then all the terms must be equal to 0

2x - 1 = 0

x = $$\dfrac{1}{2}$$

2y + 1 = 0

y = $$-\dfrac{1}{2}$$

2z + 1 = 0

z = $$-\dfrac{1}{2}$$

Substituting the values in equation 2, we get,

$$4\left(\dfrac{1}{2}\ -\ \left(-\dfrac{1}{2}\right)\ -\ \left(-\dfrac{1}{2}\right)\right)\ =\ 3\ +\ a$$

$$4\left(\dfrac{3}{2}\right)\ =\ 3\ +\ a$$

$$6\ =\ 3\ +\ a$$

$$\ a\ =\ 3$$

Therefore, the correct answer is option A.

When given more than one equations, stating the fact that there is a common root, 
We need to equate the two equations to get discernible values for $$x$$

Here, we are given three equations with the values of $$m$$, $$n$$

$$x^2+mx+9=x^2+\left(m+n\right)x+35$$
$$mx+9=mx+nx+35$$
$$nx=-26$$

Similarly, we can do it for the other equation as well, 
$$x^2+nx+17=x^2+\left(m+n\right)x+35$$
$$mx=-18$$

Substituting the value of either $$mx$$ or $$nx$$ in the original equations, we get 
$$x^2-18+9=0$$
$$x^2=9$$
$$x=\pm\ 3$$
Since we are given that the root is negative, $$x=-3$$

$$n=-\dfrac{26}{-3}$$
$$m=-\dfrac{18}{-3}$$

$$3n=26$$
$$2m=12$$

$$2m+3n=38$$

From the sum and product of roots, we get: $$\alpha\ +\beta\ =-\dfrac{\lambda}{3}$$ and $$\alpha\ \beta\ =-\dfrac{1}{3}$$

Simplifying the expression given in the question, we get: $$\dfrac{\alpha^2+\beta^2\ }{\alpha^2\beta^2\ }=15$$
and substituting the denominator's value as 1/9, we get:$$\alpha^2+\beta^2\ =\dfrac{15}{9}$$

We want the expression $$\alpha^3+\beta^3\ $$, so multiplying both sides by $$\alpha+\beta$$, we get:
$$\alpha^3+\beta^3+\alpha\beta\left(a+\beta\ \right)=\dfrac{15}{9}\left(\alpha\ +\beta\ \right)$$
$$\alpha^3+\beta^3+\dfrac{\lambda}{9}\ =\dfrac{15}{9}\left(-\dfrac{\lambda}{3}\ \right)$$
$$\alpha^3+\beta^3+\dfrac{\lambda}{9}\ =-\dfrac{5\lambda}{9}-\dfrac{\lambda}{9}=-\dfrac{2\lambda\ }{3}\ \ $$

We would still need to find the value of $$\lambda$$
This we can do from the initial relation we had:

$$\alpha^2+\beta^2\ =\dfrac{15}{9}$$
$$\alpha^2+\beta^2\ =\left(\alpha+\beta\right)^2-2\alpha\ \beta\ \ \ =\dfrac{15}{9}$$
$$\dfrac{\lambda^2}{9}+\frac{2}{3}\ \ \ =\dfrac{15}{9}$$
$$\dfrac{\lambda^2}{9}\ =\dfrac{15-6}{9}=\dfrac{9}{9}=1$$
This would finally give us $$\lambda^2=9$$

Using this in our required expression, we get:

$$\left(\alpha^3+\beta^3\right)^2=\left(-\dfrac{2\lambda}{3}\ \ \right)^2=\dfrac{4\times\ 9}{9}=4$$

Therefore, Option B is the correct answer. 

In such questions, we should be trying to complete the squares. 
We see a $$xy$$ term; we need to accommodate that in a square that has both x and y terms. 

Since there is only one other term with x, we also need to have it entirely in the square. 
$$\left(2x-y\right)^{^2}=4x^2+y^2-4xy$$
Using this in the given equation, we are left with $$\left(2x-y\right)^{^2}+3y^2+3-6y$$

This can be written as $$\left(2x-y\right)^{^2}+3\left(y^2+1-2y\right)$$
$$\left(2x-y\right)^{^2}+3\left(y-1\right)^2=0$$

Since both the squares add up to 0, this is only possible when the squares themselves are 0
This would give us y=1 from the second term, and using that, we get x= 1/2 from the first term. 

Therefore the value of 4x+5y will be 2+5 = 7

Hence, 7 is the correct answer. 

It is given that $$2^{4x^{2}}-2^{2x^{2}+x+16}+2^{2x+30}=0$$, which can be written as:

=>$$\left(2^{2x^2}\right)^2-2^{2x^2}\cdot2^{x+15}\cdot2^1+\left(2^{x+15}\right)^{^2}=0$$

=> $$\left(2^{2x^2}-2^{x+15}\right)^{^2}=0$$

=> $$2^{2x^2}-2^{x+15}=0$$ (Since $$\left(a-b\right)^2\ =\ 0\ =>\ a-b\ =0$$)

=> $$2x^2\ =\ x+15$$

=> $$2x^2-x-15=0$$ 

=> $$2x^2-6x+5x-15=0$$

=> $$2x\left(x-3\right)+5\left(x-3\right)=0$$

=> $$\left(2x+5\right)\left(x-3\right)\ =\ 0$$

Hence, the possible values of x are $$-\frac{5}{2}$$, and $$3$$, respectively.

Therefore, the sum of the possible values is $$\left(3-\frac{5}{2}\right)=\frac{1}{2}$$

The correct option is D

Given that -2 is a root of the given cubic equation.

=> Dividing the given equation by (x + 2), Using the Horners method of synthetic division:

coefficient of $$x^2$$ is 1, and coefficient of x is (2r+1)-2 = 2r-1 and the constant term = (4r-1)-2(2r-1) = 1.

=> The quadratic obtained by dividing the cubic = $$x^2+\left(2r-1\right)x+1=0$$, Since, this equation has 2 real roots => Discriminant should be greater than 0

=> $$\left(2r-1\right)^2>4$$ => 2r-1 > 2 or 2r-1 < -2 => r > 3/2 or r < -1/2.

=> Minimum possible non-negative integer value of r is 2.

It is given that $$x^2 + bx + c = 0$$ has two real roots. Let the roots of the equation be $$\alpha\ ,\beta\ $$. ($$\alpha\ \ >\ \beta\ $$)

Then, we can say that $$\frac{1}{\alpha\ }-\frac{1}{\beta\ }=\frac{1}{3}$$ .... Eq(1)

Similarly, $$\frac{1}{\alpha\ ^2}+\frac{1}{\beta^2\ }=\frac{5}{9}$$ .... Eq (2)

Eq(2) can be written as $$\left(\frac{1}{\alpha\ }-\frac{1}{\beta\ }\right)^{^2}+2\cdot\frac{1}{\alpha\ }\cdot\frac{1}{\beta\ }=\frac{5}{9}$$

=> $$\left(\frac{1}{3}\right)^{^2}+2\cdot\frac{1}{\alpha\ }\cdot\frac{1}{\beta\ }=\frac{5}{9}$$

=> $$\frac{2}{\alpha\ \cdot\beta\ }=\frac{4}{9}=>\ \frac{1}{\alpha\ \cdot\beta\ }=\frac{2}{9}$$

=> $$\alpha\ \cdot\beta\ =\frac{9}{2}$$

We know that the product of the roots is equal to c, which implies$$c=\frac{9}{2}$$

We also know that the sum of the roots is equal to -b. 

=> $$\frac{1}{\alpha\ ^2}+\frac{1}{\beta\ ^2}=\left(\frac{1}{\alpha\ }+\frac{1}{\beta\ }\right)^{^2}-\frac{2}{\alpha\ \beta\ }=\frac{5}{9}$$

=> $$\left(\ \frac{\ \alpha\ +\beta\ }{\alpha\ \beta\ }\right)^{^2}-\frac{4}{9}=\frac{5}{9}$$

=> $$\left(\ \frac{\ \alpha\ +\beta\ }{\alpha\ \beta\ }\right)^{^2}=\left(1\right)^2$$

=> $$\alpha\ +\beta\ \ =\ \pm\ \alpha\ \beta\ $$

Hence, the maximum value of b is $$\frac{9}{2}$$.

Hence, the maximum value of (b+c) is 9

Given a and b are the distinct roots of the equation $$2x^{2} - 6x + k = 0$$

=> a + b = -(-6/2) = 3 (Sum of the roots)

=> ab = k/2 (Product of the roots)

Now, (a+b) and ab are the roots of the quadratic equation $$x^{2} + px + p = 0$$

=> a + b + ab = -p => 3 + k/2 = -p ---(1)

=> (a + b)(ab) = p => 3(k/2) = p ---(2)

$$3+\dfrac{k}{2}=-\dfrac{3k}{2}$$ => 2k = -3 => k = $$-\dfrac{3}{2}$$

p = $$\dfrac{3k}{2}=\dfrac{3}{2}\left(-\dfrac{3}{2}\right)=-\dfrac{9}{4}$$

=> 8(k-p) = $$8\left(-\frac{3}{2}+\frac{9}{4}\right)=-12+18=6$$

It is given that $$\left(x-1\right)^2+2kx+11=0$$ has no real roots. (Where k is the largest integer)

$$\left(x-1\right)^2+2kx+11=0$$, which can be written as:

=> $$x^2-2x+1+2kx+11=0$$

=> $$x^2+2\left(k-1\right)x+12=0$$

We know that for no real roots, D < 0 => b^2 -4ac < 0

Hence, $$\left\{2\left(k-1\right)\right\}^2-4\cdot1\cdot12\ <0$$

=> $$4\left(k-1\right)^2<48$$

=> $$\left(k-1\right)^2<12$$

Since k is an integer, it implies (k-1) is also an integer. 

Therefore, from the above inequality, we can say that the largest possible value of (k-1) = 3 => The largest possible value of k is 4.

Now we need to calculate the least possible value of $$\frac{k}{4y}+9y$$. 

$$\frac{k}{4y}+9y$$ can be written as $$\frac{4}{4y}+9y\ =\ \frac{1}{y}+9y$$

The least possible value of $$9y\ +\frac{1}{y}$$ can be calculated using A.M-G.M inequality.

Using A.M-G.M inequality, we get:

$$\ \frac{\ 9y+\frac{1}{y}}{2}\ge\ \sqrt{\ 9y\times\ \frac{1}{y}}$$

=> $$\ \frac{\ 9y+\frac{1}{y}}{2}\ge\ \sqrt{\ 9}$$

=> $$\ \frac{\ 9y+\frac{1}{y}}{2}\ge3$$

=> $$\ \ 9y+\frac{1}{y}\ge6$$

Hence, the least possible value is 6

$$2x^{2}+kx+5=0$$ has no real roots so D<0

$$k^2-40\ <0$$

$$\left(k-\sqrt{40}\right)\left(k+\sqrt{40}\right)<0$$

$$k\in\left(-\sqrt{40},\sqrt{40}\right)$$

$$x^{2}+(k-5)x+1=0$$ has two distinct real roots so D>0

$$\left(k-5\right)^2-4>0$$

$$k^2-10k+21>0$$

$$\left(k-3\right)\left(k-7\right)>0$$

$$k\in\left(-\infty\ ,3\right)∪\left(7,\infty\ \right)$$

Therefore possibe value of k are -6, -5, -4, -3, -2, -1, 0, 1, 2

In 9 total 9 integer values of k are possible.

a, b, c, m and n are integers so if one root is $$3+2\sqrt{2}$$ then the other root is $$3-2\sqrt{2}$$

Sum of roots = 6 = -b/a  or b= -6a

Product of roots = 1 = c/a  or c=a

a, b, c, m and n are integers so if one root is $$4+2\sqrt{3}$$ then the other root is $$4-2\sqrt{3}$$

Sum of roots = 8 = -m/a or m = -8a

product of roots = 4 = n/a or n = 4a

$$(\frac{b}{m}+\frac{c-2b}{n})$$

= $$\frac{6a}{8a}+\frac{\left(a+12a\right)}{4a}=\frac{3}{4}+\frac{13}{4}=\frac{16}{4}=4$$

Let the roots of the given equation $$5x^{3} + cx^{2} - 10x + 9 = 0$$ be r, -r and p

r - r + p = $$-\frac{c}{5}$$

p = $$-\frac{c}{5}$$ ...... (1)

$$-r^2-pr+pr=-2$$

$$r^2=2$$ ...... (2)

$$-r^2p=-\frac{9}{5}$$

$$p=\frac{9}{10}$$ ...... (3)

Substituting p in (1), we get

$$\frac{9}{10}=-\frac{c}{5}$$

$$-\frac{9}{2}=c$$

The answer is option A.

$$b^2 < 4ac$$ means that the discriminant is less than 0. Therefore, f(x)>0 for all x if the coefficient of $$x^2$$ is positive, and f(x)<0 for all x if the coefficient of $$x^2$$ is negative.

We are given that f(m)<0 and m is an integer.

So the set containing values of m will either be empty if the coefficient of $$x^2$$ is positive, or it will be a set of all integers if the coefficient of $$x^2$$ is negative.

Let $$\frac{x^2-6x+10}{3-x}=p$$

$$x^2-6x+10=3p-px$$

$$x^2-\left(6-p\right)x+10-3p=0$$

Since the equation will have real roots,

$$\left(6-p\right)^2-4\times\left(10-3p\right)\ge0$$

$$p^2-12p+12p+36-40\ge0$$

$$p^2\ge4$$

$$p\ge2\ ,\ p\le-2$$

Now, when $$p=-2$$, $$x = 4$$.  Since it is given that $$x<3$$, thus this value will be discarded.

Now, $$\frac{1}{2}$$ and $$-\frac{1}{2}$$ do not come in the mentioned range.

when $$p=2$$, $$x = 2$$

Thus, the minimum possible value of p will be 2.

Thus, the correct option is B.

Alternate explanation:

Since $$x<3$$,

$$3-x>0$$

Let $$3-x=y$$. So, $$y>0$$.

Now, $$\frac{x^2-6x+10}{3-x}=\frac{x^2-6x+9+1}{3-x}$$

=> $$\frac{\left(3-x\right)^2+1}{3-x}$$

Since $$3-x=y$$, the equation will transform to $$\frac{y^2+1}{y}$$ or $$y+\frac{1}{y}$$

The minimum value of the expression $$y+\frac{1}{y}$$ for $$y>0$$ will at $$y=1$$

i.e., Minimum value = $$1+1=2$$

Thus, the correct option is B.

Given a, b, c are rational numbers.

Hence a, b, c are three numbers that can be written in the form of p/q.

Hence if one both the root is 2+$$\sqrt{\ 3}$$ and considering the other root to be x.

The sum of the roots and the product of the two roots must be rational numbers.

For this to happen the other root must be the conjugate of $$2+\sqrt{\ 3}$$ so the sum and the product of the roots are rational numbers which are represented by: $$-\frac{b}{a},\ \frac{c}{a}$$

Since the quadratic equation has negative b, it will cancel the negative sign of the sum of the roots. Hence, the sum of the roots and the products of the roots will be represented by  $$-\frac{b}{a},\ \frac{c}{a}$$

Hence the sum of the roots is 2+$$\sqrt{\ 3}+2-\sqrt{\ 3}$$ = 4.

The product of the roots is $$\left(2+\sqrt{\ 3}\right)\cdot\left(2-\sqrt{\ 3}\right)\ =\ 1$$

b/a = 4, c/a = 1.
b = 4*a, c= a.
Since b = $$c^3$$
4*a = $$a^3$$
$$a^2=\ 4.$$
a = 2 or -2.
|a| = 2

$$f(x)=\frac{x^{2}+2x+4}{2x^{2}+4x+9}$$

If we closely observe the coefficients of the terms in the numerator and denominator, we see that the coefficients of the $$x^2$$ and x in the numerators are in ratios 1:2. This gives us a hint that we might need to adjust the numerator to decrease the number of variables.

$$f(x)=\frac{x^2+2x+4}{2x^2+4x+9}=\frac{x^2+2x+4.5-0.5}{2x^2+4x+9}$$

= $$\frac{x^2+2x+4.5}{2x^2+4x+9}-\frac{0.5}{2x^2+4x+9}$$

= $$\frac{1}{2}-\frac{0.5}{2x^2+4x+9}$$

Now, we only have terms of x in the denominator.

The maximum value of the expression is achieved when the quadratic expression $$2x^2+4x+9$$ achieves its highest value, that is infinity.

In that case, the second term becomes zero and the expression becomes 1/2. However, at infinity, there is always an open bracket ')'.

To obtain the minimum value, we need to find the minimum possible value of the quadratic expression.

The minimum value is obtained when 4x + 4 = 0 [d/dx = 0]

x=-1.

The expression comes as 7. 

The entire expression becomes 3/7.

Hence, $$[\frac{3}{7},\frac{1}{2})$$

Let the number of Covid patients in Hospitals A and B be x and x+21, respectively. Then, it has been given that:

$$\dfrac{200}{x}-\dfrac{152}{x+21}=3$$

$$\dfrac{\left(200x+4200-152x\right)}{x\left(x+21\right)}=3$$

$$\dfrac{\left(48x+4200\right)}{x\left(x+21\right)}=3$$

$$16x+1400=x\left(x+21\right)$$

$$x^2+5x-1400=0$$

(x+40)(x-35)=0

Hence, x=35.

We have :
$$x^2-xy-x\ =22\ \ \ \ \ \ \left(1\right)$$ 
And $$y^2-xy+y\ =34\ \ \ \ \ \ \left(2\right)$$         
Adding (1) and (2)
we get $$x^2-2xy+y^2-x+y\ =56$$
we get $$\left(x-y\right)^2-\ \left(x-y\right)\ =56$$
Let (x-y) =t
we get $$t^2-t=56$$
$$t^2-t-56=0$$
(t-8)(t+7) =0
so t=8
so x-y =8

The quadratic equation of the form $$\mid x^2 - 4x - 13 \mid = r$$ has its minimum value at x = -b/2a, and hence does not vary irrespective of the value of x.

Hence at x = 2 the quadratic equation has its minimum.

Considering the quadratic part : $$\left|x^2-4\cdot x-13\right|$$. as per the given condition, this must-have 3 real roots.

The curve ABCDE represents the function $$\left|x^2-4\cdot x-13\right|$$. Because of the modulus function, the representation of the quadratic equation becomes :

ABC'DE. 

There must exist a value, r such that there must exactly be 3 roots for the function. If r = 0 there will only be 2 roots, similarly for other values there will either be 2 or 4 roots unless at the point C'.

The point C' is a reflection of C about the x-axis. r is the y coordinate of the point C' :

The point C which is the value of the function at x = 2, = $$2^2-8-13$$

= -17, the reflection about the x-axis is 17.

Alternatively,

$$\mid x^2 - 4x - 13 \mid = r$$ .

This can represented in two parts :

$$x^2-4x-13\ =\ r\ if\ r\ is\ positive.$$

$$x^2-4x-13\ =\ -r\ if\ r\ is\ negative.$$

Considering the first case : $$x^2-4x-13\ =r$$

The quadraticequation becomes : $$x^2-4x-13-r\ =\ 0$$

The discriminant for this function is : $$b^2-4ac\ =\ 16-\ \left(4\cdot\left(-13-r\right)\right)=68+4r$$

SInce r is positive the discriminant is always greater than 0 this must have two distinct roots.

For the second case :

$$x^2-4x-13+r\ =\ 0$$ the function inside the modulus is negaitve

The discriminant is $$16\ -\ \left(4\cdot\left(r-13\right)\right)\ =\ 68-4r$$

In order to have a total of 3 roots, the discriminant must be equal to zero for this quadratic equation to have a total of 3 roots.

Hence $$\ 68-4r\ =\ 0$$

r = 17, for r = 17 we can have exactly 3 roots.

To have real roots the discriminant should be greater than or equal to 0.

So, $$m^2-8n\ge0\ \&\ 4n^2-4m\ge0$$

=> $$m^2\ge8n\ \&\ n^2\ge m$$

Since m,n are positive integers the value of m+n will be minimum when m=4 and n=2.

.'. m+n=6.

$$(x^{2}-7x+11)^{(x^{2}-13x+42)}=1$$

if $$(x^{2}-13x+42)$$=0 or $$(x^{2}-7x+11)$$=1 or $$(x^{2}-7x+11)$$=-1 and $$(x^{2}-13x+42)$$ is even number

For x=6,7 the value $$(x^{2}-13x+42)$$=0

$$(x^{2}-7x+11)$$=1 for x=5,2.

$$(x^{2}-7x+11)$$=-1 for x=3,4 and for X=3 or 4, $$(x^{2}-13x+42)$$ is even number.

.'. {2,3,4,5,6,7} is the solution set of x.

.'. x can take six values. 

Let $$a=x+\frac{1}{x}$$
So, the given equation is $$a^2-3a+2=0$$
So, $$a$$ can be either 2 or 1.

If $$a=1$$, $$x+\frac{1}{x}=1$$ and it has no real roots. 
If $$a=2$$, $$x+\frac{1}{x}=2$$ and it has exactly one real root which is $$x=1$$

So, the total number of distinct real roots of the given equation is 1

$$\left(x^2-5x+7\right)^{x+1}=1$$

There can be a solution when $$\left(x^2-5x+7\right)=1$$ or $$x^2-5x\ +6=0$$

or x=3 and x=2

There can also be a solution when x+1 = 0 or x=-1

Hence three possible solutions exist.

We have, $$\mid x^2 - x - 6 \mid = x + 2$$

=> |(x-3)(x+2)|=x+2

For x<-2, (3-x)(-x-2)=x+2

=> x-3=1   =>x=4 (Rejected as x<-2)

For -2$$\le\ $$x<3, (3-x)(x+2)=x+2    =>x=2,-2

For x$$\ge\ $$3, (x-3)(x+2)=x+2   =>x=4

Hence the product =4*-2*2=-16

The roots of $$x^2 - 4x - log_{2}{A} = 0$$ will be real and distinct if and only if the discriminant is greater than zero

16+4*$$log_{2}{A}$$ > 0

$$log_{2}{A}$$ > -4

A> 1/16

Given,

The quadratic equation $$x^2 + bx + c = 0$$ has two roots 4a and 3a

7a=-b

12$$a^2$$ = c

We have to find the value of  $$b^2 + c = 49a^2+ 12a^2=61a^2$$

Now lets verify the options 

61$$a^2$$ = 3721 ==> a= 7.8 which is not an integer

61$$a^2$$ = 361 ==> a= 2.42 which is not an integer

61$$a^2$$ = 427 ==> a= 2.64 which is not an integer

61$$a^2$$ = 549 ==> a= 3 which is an integer

For x <0, -x($$6x^2+1$$) = $$5x^2$$

=> ($$6x^2+1$$) = -5x

=> ($$6x^2 + 5x+ 1$$) = 0 

=>($$6x^2 + 3x+2x+ 1$$) = 0

=> (3x+1)(2x+1)=0    =>x=$$\ -\frac{\ 1}{3}$$  or x=$$\ -\frac{\ 1}{2}$$

For x=0, LHS=RHS=0    (Hence, 1 solution)

For x >0, x($$6x^2+1$$) = $$5x^2$$

=> ($$6x^2 - 5x+ 1$$) = 0 

=>(3x-1)(2x-1)=0    =>x=$$\ \frac{\ 1}{3}$$   or   x=$$\ \frac{\ 1}{2}$$

Hence, the total number of solutions = 5

Given that $$U^{2}+(U-2V-1)^{2}$$= −$$4V(U+V)$$ 

$$\Rightarrow$$ $$U^{2}+(U-2V-1)(U-2V-1)$$= −$$4V(U+V)$$

$$\Rightarrow$$ $$U^{2}+(U^2-2UV-U-2UV+4V^2+2V-U+2V+1)$$ = −$$4V(U+V)$$ 

$$\Rightarrow$$ $$U^{2}+(U^2-4UV-2U+4V^2+4V+1)$$ = −$$4V(U+V)$$ 

$$\Rightarrow$$ $$2U^2-4UV-2U+4V^2+4V+1=−4UV-4V^2$$

$$\Rightarrow$$ $$2U^2-2U+8V^2+4V+1=0$$

$$\Rightarrow$$ $$2[U^2-U+\dfrac{1}{4}]+8[V^2+\dfrac{V}{2}+\dfrac{1}{16}]=0$$

$$\Rightarrow$$ $$2(U-\dfrac{1}{2})^2+8(V+\dfrac{1}{4})^2=0$$

Sum of two square terms is zero i.e. individual square term is equal to zero. 

$$U-\dfrac{1}{2}$$ = 0 and $$V+\dfrac{1}{4}$$ = 0

U = $$\dfrac{1}{2}$$ and V = $$-\dfrac{1}{4}$$

Therefore, $$U+3V$$ = $$\dfrac{1}{2}$$+$$\dfrac{-1*3}{4}$$ = $$\dfrac{-1}{4}$$. Hence, option C is the correct answer.

Let f(x) = $$2x^2−ax+2$$. We can see that f(x) is a quadratic function. 

For, f(x) > 0, Discriminant (D) < 0

$$\Rightarrow$$ $$(-a)^2-4*2*2<0$$

$$\Rightarrow$$ (a-4)(a+4)<0

$$\Rightarrow$$ a $$\epsilon$$ (-4, 4)

Therefore, integer values that 'a' can take = {-3, -2, -1, 0, 1, 2, 3}

Let g(x) = $$x^2−bx+8$$. We can see that g(x) is also a quadratic function. 

For, g(x)≥0, Discriminant (D) $$\leq$$ 0

$$\Rightarrow$$ $$(-b)^2-4*8*1<0$$

$$\Rightarrow$$ $$(b-\sqrt{32})(b+\sqrt{32})<0$$

$$\Rightarrow$$ b $$\epsilon$$ (-$$\sqrt{32}$$, $$\sqrt{32}$$)

Therefore, integer values that 'b' can take = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

We have to find out the largest possible value of $$2a−6b$$. The largest possible value will occur when 'a' is maximum and 'b' is minimum. 

a$$_{max}$$ = 3, b$$_{min}$$ = -5

Therefore, the largest possible value of $$2a−6b$$ = 2*3 - 6*(-5) = 36. 

We know that $$x^2 - x - 1=0$$
Therefore $$x^4 = (x+1)^2 = x^2+2x+1 = x+1 + 2x+1 = 3x+2$$
Therefore, $$2x^4 = 6x+4$$

We know that $$x>0$$ therefore, we can calculate the value of $$x$$ to be $$\frac{1+\sqrt{5}}{2}$$
Hence, $$2x^4 = 6x+4 = 3+3\sqrt{5}+4 = 3\sqrt{5}+7$$

b = sum of the roots taken 2 at a time.
Let the roots be n-1, n and n+1.
Therefore, $$b = (n-1)n + n(n+1) + (n+1)(n-1) = n^2 - n + n^2 + n + n^2 - 1$$
$$b = 3n^2 - 1$$. The smallest value is -1.

Let the function be $$ax^2 + bx + c$$.

We know that x=0 value is 1 so c=1.

So equation is $$ax^2 + bx + 1$$.

Now max value is 3 at x = 1.

So after substituting we get a + b = 2.

If f(x) attains a maximum at 'a' then the differential of f(x) at x=a, that is, f'(a)=0.

So in this question f'(1)=0

=> 2*(1)*a+b = 0

=> 2a+b = 0.

Solving the equations we get a=-2 and b=4.

$$ -2x^2 + 4x + 1$$ is the equation and on substituting x=10, we get -159.

From 1st equation we know that $$(x)^2 = y^2 $$

Substituting this in 2nd equation. we get , $$2*x^2 - 2*x*k + k^2-1 =0 $$ and for unique solution $$b^2-4ac=0$$ must satisfy.

This is possible only when k = $$\sqrt{2}$$

Let $$\alpha $$ be equal to k.

=> f(x) = $$x^2-(k-2)x-(k+1) = 0$$

p and q are the roots

=> p+q = k-2 and pq = -1-k

We know that $$(p+q)^2 = p^2 + q^2 + 2pq$$

=> $$ (k-2)^2 = p^2 + q^2 + 2(-1-k)$$

=> $$p^2 + q^2 = k^2 + 4 - 4k + 2 + 2k$$

=> $$p^2 + q^2 = k^2 - 2k + 6$$

This is in the form of a quadratic equation.

The coefficient of $$k^2$$ is positive. Therefore this equation has a minimum value.

We know that the minimum value occurs at x = $$\frac{-b}{2a}$$

Here a = 1, b = -2 and c = 6

=> Minimum value occurs at k = $$\frac{2}{2}$$ = 1

If we substitute k = 1 in $$k^2-2k+6$$, we get 1 -2 + 6 = 5.

Hence 5 is the minimum value that $$p^2+q^2$$ can attain.

The given equation can be written as : $$A^2 * (x-1) + B^2 * x = x^2 - x$$
=> $$x^2 + x(-1 - A^2 - B^2) + A^2 = 0$$
Discriminant of the equation = $$ (-1 - A^2 - B^2) ^2 - 4A^2$$
=$$ A^4 + B^4 + 1 - 2A^2 + 2B^2 + 2A^2B^2$$
= $$ A^4 + B^4 + 1 - 2A^2 - 2B^2 + 2A^2B^2 + 4B^2$$
= $$ (A^2 + B^2 - 1)^2 + 4B^2$$
>= 0, 0 when B =0 and A =1
Hence, the number of roots can be 1 or 2.
Option d) is the correct answer.

Let the optimum number of samosas be 200+20n

So, price of each samosa = (2-0.1*n)

Total price of all samosas = (2-0.1*n)*(200+20n) = $$400 - 20n + 40n - 2n^2$$ = $$400 + 20n - 2n^2$$

This quadratic equation attains a maximum at n = -20/2*(-2) = 5

So, the number of samosas to get the maximum revenue = 200 + 20*5 = 300

We know that quadratic equation can be written as $$x^2$$-(sum of roots)*x+(product of the roots)=0.
Ujakar ended up with the roots (4, 3) so the equation is $$x^2$$-(7)*x+(12)=0 where the constant term is wrong.

Keshab got the roots as (3, 2) so the equation is $$x^2$$-(5)*x+(6)=0 where the coefficient of x is wrong .

So the correct equation is $$x^2$$-(7)*x+(6)=0. The roots of above equations are (6,1).

It can be clearly seen that if b=1 then $$x^2(x - a) + (x - a) = 0$$ an the equation gives only 1 real value of x

$$x_1 + x_2 = 2$$
and $$7x_2 - 4x_1 = 47$$
So $$x_1 = -3$$ and $$x_2 = 5$$ 
And $$c = x_1 \times x_2 = -15$$

For summation of square of roots to be zero, individual roots should be zero.
Hence summation should be zero i.e.  A-3=0 ; A = 3
And product of roots will also be zero i.e. A-2 = 0 ; A =2
So there is no unique value of A which can satisfy above equation.

$$\frac{(1-d^3)}{(1-d)}$$ = $$1+d^2+d$$ (where $$d \neq 1$$)
Let's say $$f(d)=1+d^2+d$$ 
Now $$f(d)$$ will always be greater than 0 and have its minimum value at d = -0.5. The value is $$\frac{3}{4}$$.
For $$d<-1$$ ; $$f(d) >1$$
$$-1<d<0$$ ; $$\frac{3}{4} <f(d)<1$$
$$0<d<1$$ ; $$1<f(d)<3$$
$$d>1$$ ; $$f(d)>3$$

So, for d > 1, f(d) > 3. Option b) is the correct answer.

If roots of given equation are reciprocal to each other than product of roots should be equal to 1.
i.e. $$\frac{6}{a} = 1$$
hence a=6

$$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$$
as $$xy+yz+xz = 0$$
so equation will be resolved to $$x^2 + y^2 + z^2$$