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

Calculation:

Given,

From the previous result, the roots of the original equation are

\( \alpha = 1 \) and \( \beta = -2 \)

We seek the quadratic whose roots are

\( \alpha + 1 = 1 + 1 = 2\) and \( \beta + 1 = -2 + 1 = -1 \)

For a quadratic with roots \(r_1\) and \(r_2\), the monic form is:

\( x^2 - (r_1 + r_2)\,x + (r_1\,r_2) = 0. \)

Here,

\( r_1 + r_2 = 2 + (-1) = 1, \)

\( r_1\,r_2 = 2 \times (-1) = -2. \)

Substituting gives:

\( x^2 - 1\cdot x + (-2) = 0 \)

\( \boxed{x^2 - x - 2 = 0} \)

Hence, the correct anwer is Option 2.

Calculation:

Given,

α and β are the roots of the quadratic equation

\(x^2 + \sqrt{\alpha}\,x + \beta = 0 \)

By Viète’s relations,

Sum of roots: \( \alpha + \beta = -\sqrt{\alpha} \),

Product of roots: \(\alpha\beta = \beta \)

From \(\alpha\beta = \beta \), either \(\beta = 0 \) or \(\alpha = 1 \).

Since β≠0 in the given options, we take \(\alpha = 1 \).

Substitute into the sum relation:

\(1 + \beta = -\sqrt{1} = -1\) ⟹ \(\beta = -2 \)

Therefore, \(\alpha = 1 \) and \(\beta = -2 \)

Hence, the correct answer is Option 4. 

Calculation:

We are given that k is a root of $x{}^2-4x+1=0$

Solving the quadratic equation:

\(x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} \)

⇒ \(x = 2 \pm \sqrt{3} \)

Thus, the two possible values of k are

⇒ \(k = 2 + \sqrt{3} \quad \text{or} \quad k = 2 - \sqrt{3} \)

We need to find  \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) \)

Using the identity for the sum of inverse tangents

\(\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \)

⇒ \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - k \cdot \frac{1}{k}}\right) \)

\(= \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - 1}\right) = \tan^{-1} \left(\frac{k + \frac{1}{k}}{0}\right) \)

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

⇒ \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \frac{\pi}{2}\)

Hence, the correct answer is Option 4.

Calculation:

Given,

The equation is \( x^2 - x + 1 = 0 \)

We need to find the value of the following expression:

\( \left( x - \frac{1}{x} \right)^2 + \left( x - \frac{1}{x} \right)^4 + \left( x - \frac{1}{x} \right)^8 \)

The equation \( x^2 - x + 1 = 0 \) is solved as follows:

\( x = \frac{1 \pm \sqrt{-3}}{2} = e^{i \pi / 3} \quad \text{or} \quad x = e^{-i \pi / 3} \)

Now, substitute the value of \( x \) into the expression \( x - \frac{1}{x} \):

\( x - \frac{1}{x} = i\sqrt{3} \)

Evaluate the powers of \( x - \frac{1}{x} \)

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

\( \left( x - \frac{1}{x} \right)^2 = (i \sqrt{3})^2 = -3 \)

\( \left( x - \frac{1}{x} \right)^4 = (-3)^2 = 9 \)

\( \left( x - \frac{1}{x} \right)^8 = 9^2 = 81 \)

Now, sum the values:

\( -3 + 9 + 81 = 87 \)

∴ 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 

\((\alpha + \beta )^2 = (\alpha - \beta)^2 - 4\alpha\beta \)

⇒ 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 = \(\rm -\dfrac{B}{A}\) and αβ = \(\rm \dfrac{C}{A}\).

Calculation:

Let's say that α and β are the two roots of the quadratic equation ax2 + bx + c = 0, then α + β = \(\rm -\dfrac{b}{a}\) and αβ = \(\rm \dfrac{c}{a}\).

It is given that α = -β.

∴ -β + β = \(\rm -\dfrac{b}{a}\)

⇒ \(\rm -\dfrac{b}{a}\) = 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: \({\rm{α }} + {\rm{β }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\) 

The product of the roots is given by:

 \({\rm{α β }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)

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, 

\(\rm\frac {1}{x-3} = \frac {1}{x + 2} - \frac 1 2\\ \Rightarrow \rm\frac {1}{x-3}= \frac{2-x-2}{2 x+4} \\ \Rightarrow\frac{-x}{2 x+4}=\frac{1}{x-3} \\ \Rightarrow\frac{1}{x-3}+\frac{x}{2 x+4}=0 \\ \Rightarrow\frac{2 x+4+x^{2}-3 x}{(x-3)(2 x+4)}=0 \\ \Rightarrow x^{2}-x+4=0 \)

∴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: \({\rm{α }} + {\rm{β }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\) 

The product of the roots is given by:  \({\rm{α β }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)

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

The sum of the roots of a quadratic equation is given by: \({\rm{α }} + {\rm{β }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\) 

The product of the roots is given by:  \({\rm{α β }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)

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 = \( - \frac{{\rm{b}}}{{\rm{a}}} \)

Product of roots  = α β = \(\frac{{\rm{c}}}{{\rm{a}}}\)

Now, 

α - β = 1

squaring both sides, we get

⇒ (α - β)² = 1²

⇒ (α + β)² - 4α β = 1

⇒ \(\rm (\frac{-b}{a})^{2} - \frac{4c}{a} = 1\)

⇒ 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,\(\frac{α ^3+β ^3}{α ^{-3}+β ^{-3}}\)= \(\frac{α ^3+β ^3}{\frac {1}{α ^{3}}+\frac{1}{β ^{3}}}\)

= \(\frac{α ^3+β ^3}{\frac {(α ^3+β ^3)}{α^3 β^3 }}\)

= (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.\(\rm \dfrac{\pi}{4}\)

∴ \(\rm \dfrac{\pi}{2}\)