Quadratic Equations MCQ Quiz - Objective Question with Answer for Quadratic Equations - Download Free PDF

Calculation:

We are given that k is a root of x2−4x+1=0

Solving the quadratic equation:

⇒ 

Thus, the two possible values of k are

⇒ 

We need to find  

Using the identity for the sum of inverse tangents

⇒ 

This expression results in an undefined value, but we know from properties of the inverse tangent that

⇒ 

Hence, the correct answer is Option 4.

Calculation:

Given,

The equation is

We need to find the value of the following expression:

The equation is solved as follows:

Now, substitute the value of into the expression

Evaluate the powers of

Now, let's evaluate each term in the expression:

Now, sum the values:

∴ The value of the expression is 87.

Hence, the correct answer is Option 3. 

Given:

The quadratic equation is x² - kx + k = 0.

One root exceeds the other by 2√3.

⇒ α - β = 2√3.

Also, 

Sum of roots: α + β = k

Product of roots: α × β = k

Calculation:

We know the following identity 

⇒ k² = (2√3)² - 4k 

⇒ k² - 12 - 4k = 0

⇒ k² - 6k + 2k -12 = 0

⇒ k(k - 6) + 2 ( k - 6) = 0

⇒ (k - 6) (k + 2) = 0

⇒ k = 6 and k = -2

Thus, the possible values of k are 6 and -2

Hence, the correct answer is Option 2.

Concept Used:

For quadratic equation, ax² + bx + c = 0,

α + β = -b/a and αβ = c/a

Calculation:

Given equation is (5 + √2) x2 - (4 + √5) x + (8 + 2√5) = 0

On comparing this equation by ax2 + bx + c = 0, we get

a = (5 + √2), b =  - (4 + √5) and c = (8 + 2√5)

Now, αβ = (8 + 2√5)/(5 + √2) and α + β = (4 + √5)/(5 + √2)

Now, We have to find the value of 2αβ/(α + β)

⇒ 2[(8 + 2√5)/(5 + √2)] / [(4 + √5)/(5 + √2)]

⇒ 2 [(8 + 2√5) (4 - √5)] / [(4 + √5)/(4 - √5)]

⇒ 2(32 + 8√5 - 8√5 - 10)/11

⇒ 44/11 = 4

∴ The required value of 2αβ/ (α + β) is 4.

Concept:

If α and β are the two roots of the quadratic equation Ax2 + Bx + C = 0, then α + β =  and αβ = .

Calculation:

Let's say that α and β are the two roots of the quadratic equation ax2 + bx + c = 0, then α + β =  and αβ = .

It is given that α = -β.

∴ -β + β = 

⇒  = 0

⇒ b = 0.

Concept:

Let us consider the standard form of a quadratic equation, 

ax² + bx + c =0

Let α and β be the two roots of the above quadratic equation. 

The sum of the roots of a quadratic equation is given by:  

The product of the roots is given by:

Calculation:

Given:  α and β are the roots of the equation x² - q(1 + x) - r = 0

⇒ x² - q - qx - r = 0

⇒ x² - qx - (q + r) = 0

Sum of roots =  α + β = q

Product of roots = αβ = - (q + r) = -q - r

To find: (1 + α)(1 + β) 

(1 + α)(1 + β) = 1 + α + β + αβ 

= 1 + q - q - r 

= 1 - r

Concept:

Degree is the highest power of the variable in a given polynomial

Calculation:

Here, 

∴Degree = 2

Hence, option (3) is correct. 

Concept: 

Let us consider the standard form of a quadratic equation,

ax² + bx + c =0

Let α and β be the two roots of the above quadratic equation. 

The sum of the roots of a quadratic equation is given by:  

The product of the roots is given by:  

Calculation:

Given:

α and β are the roots of the equation x² + px + q = 0

Sum of roots =  α + β = -p

Product of roots = αβ = q

We know that (a + b)² = a² + b² + 2ab

So, (α + β)² = α² + β² + 2αβ

⇒ (-p)² = α² + β² + 2q

∴ α² + β² = p² - 2q

Concept used:

If p(x) be a function and (x - a) be a factor of p(x) then, p(a) = 0

Calculation:

x + 4 is a factor of 3x² + kx + 8, so x = -4 will be a solution of this equation

⇒ 3(-4)² + k(-4) + 8 = 0

⇒ 4k = 48 + 8

Concept:

Let us consider the standard form of a quadratic equation, ax² + bx + c =0

Let α and β be the two roots of the above quadratic equation. 

The sum of the roots of a quadratic equation is given by:  

The product of the roots is given by:  

Calculation:

Given: difference between the roots of ax² + bx + c = 0 is 1

Let α and β be the two roots of the above quadratic equation. 

Sum of roots = α + β = 

Product of roots  = α β = 

Now, 

α - β = 1

squaring both sides, we get

⇒ (α - β)² = 1²

⇒ (α + β)² - 4α β = 1

⇒ 

⇒ b² - 4ac = a²

⇒ b² = a² + 4ac

∴ b² = a(a + 4c)

From the given equation, a = 1, b = k, c = k

For repeated roots, b² – 4ac = 0

⇒ k² – 4k = 0

⇒ k(k – 4) = 0

∴ k = 4 or k = 0.

Thus, the correct answer is option 3.

Concept:

Consider a quadratic equation: ax2 + bx + c = 0.

Let, α and β are the roots.

• Sum of roots = α + β = -b/a
• Product of the roots = α × β = c/a
   

Calculation:

Given quadratic equation:  3x2 + 57x - 5 = 0

Let  α and β are roots, then 

α + β = -57/3,  αβ = -5/3

Now,= 

= 

= (α β)³

= (-5/3)³

= -125/27

Hence, option (4) is correct.

Concept:

The modulus value is not negative.

tan^-1 (- x) = - tan^-1 (x)

Calculations:

 Given, equation is  x² + 5|x| - 6 = 0 

⇒|x²| + 5|x| - 6 = 0 

⇒|x²| + 6|x| - |x| - 6 = 0

⇒|x| (|x|+ 6) - 1 (|x| + 6) = 0

⇒ (|x| + 6) (|x| - 1)= 0

⇒(|x| + 6) = 0  and (|x| - 1) = 0

⇒ |x| = - 6  and |x| = 1

But |x| = - 6  which is not possible because value of modulus is not negative.

⇒ |x| = 1

⇒ x = 1 and x = -1

Given , α and β are toots of the equation x² + 5|x| - 6 = 0 

Hence, α = 1 and β = -1.

Now, consider, |tan-1 α - tan-1 β| = |tan-1 (1) - tan-1 (- 1)|

⇒ |tan-1 (1) + tan-1 (1)|

⇒ |2 tan-1 (1)|

⇒ 2.

∴