Test: Quadratic Equations- 1 - CAT MCQ

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 + √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 + √3 so the product and the sum of the roots are rational numbers which are represented by: b/a, c/a

Hence the sum of the roots is 2 + √3 + 2 - √3 = 4.

The product of the roots is (2 + √3)⋅(2 − √3) = 1 

b/a = 4, c/a = 1.

b = 4*a, c= a.

Since b = c³ 

4*a = a³ 

a² = 4.

a = 2 or -2.

|a| = 2

x₁ + x₂ = 2

and 7x₂ - 4x₁ = 47

So x₁ = -3 and x₂ = 5 

And c = x₁ x  x₂ = -15

Let α be equal to k.

⇒ f(x) = x² − (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 = p² + q² + 2pq

⇒ (k − 2)² = p² + q² + 2(−1 − k)

⇒ p² + q² = k² + 4 − 4k + 2 + 2k

⇒ p² + q² = k² − 2k + 6

This is in the form of a quadratic equation.

The coefficient of k² is positive. Therefore this equation has a minimum value.

We know that the minimum value occurs at x = -b/2a

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

⇒ Minimum value occurs at k = 2/2 = 1

If we substitute k = 1 in k² − 2k + 6, we get 1 -2 + 6 = 5.

Hence 5 is the minimum value that p² + q² can attain.

Let the function be ax² + bx + c.

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

So equation is ax² + 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, fʼ(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.

-2a² + 4x + 1 is the equation and on substituting x=10, we get -159.

2x² + kx + 5 = 0 has no real roots so D < 0

k² −40 < 0

x² + (k − 5)x + 1 = 0 has two distinct real roots so D > 0

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√2 then the other root is 3 - 2√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√3 then the other root is 4 - 2√3

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

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

It is given that which can be written as:

Hence, the possible values of x are -5/2 and 3, respectively.

Therefore, the sum of the possible values is (3 - 5/2) = 1/2

The quadratic equation of the form ∣x² − 4x − 13∣ = 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 : |x² − 4 ⋅ x − 13|. as per the given condition, this must-have 3 real roots.

The curve ABCDE represents the function |x² − 4 ⋅ x − 13|. 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, = 2² − 8 − 13

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

Alternatively,

|x² - 4x - 13| = r

This can represented in two parts :

Considering the first case : x² − 4x − 13 = r

The quadraticequation becomes : x² − 4x − 13 − r = 0

The discriminant for this function is : b² − 4ac = 16− (4⋅(−13 − r)) = 68 + 4r

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

For the second case :

x² − 4x − 13 + r = 0 the function inside the modulus is negaitve

The discriminant is 16 − (4⋅(r−13)) = 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.

(1 - d³)/(1-d) = 1 + d² + d (where d ≠ 1)

Let's say f (d) = 1 + d² + d

Now f(d) will always be greater than 0 and have its minimum value at d = -0.5. The value is 3/4.

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

Given,

The quadratic equation x² + bx + c = 0 has two roots 4a and 3a

7a = -b

12a² = c

We have to find the value of b² + c = 49a² + 12a² = 61a² 

Now lets verify the options 

61a² = 3721 ⇒ a= 7.8 which is not an integer

61a² = 361 ⇒ a= 2.42 which is not an integer

61a² = 427 ⇒ a= 2.64 which is not an integer

61a² = 549 ⇒ a= 3 which is an integer