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

In the given solution, the student attempted to complete the square by adding 9 on both sides of the equation. However, this step is incorrect. The correct algebraic identity for completing the square is adding the square of half the coefficient of the linear term.

Key Points

•  In the given quadratic equation, the student correctly completed the square and arrived at the equation:

(x + 3)^2 = 36

• To solve for x, the student correctly took the square root of both sides, yielding:

x + 3 = ±6

• However, when solving for x, the student only considered the positive value of 6 and subtracted 3 to find:

x = 6 - 3 = 3

• This represents the positive root of the quadratic equation.
• However, the quadratic equation can have two solutions, one positive and one negative, unless it is a perfect square. In this case, the student neglected to consider the negative root.
• To include the negative root, we would subtract 6 from -3:
• x = -3 - 6 = -9
• Thus, the student's error lies in omitting the negative root (-9) as a valid solution to the quadratic equation. It's important to consider both the positive and negative roots when solving quadratic equations, unless specified otherwise or when it is evident that the equation only has one root.

Concept:

• Quadratic Equation: An equation of the form ax² + bx + c = 0 where a, b, and c are real constants, and a ≠ 0.
• Discriminant (D): It is used to determine the nature of the roots.
  D = b² - 4ac
• If D > 0: Roots are real and distinct.
• If D = 0: Roots are real and equal.
• If D < 0: Roots are complex (imaginary).
• For both roots to be positive: Sum of roots > 0 and Product of roots > 0
• For both roots to be negative: Sum of roots < 0 and Product of roots > 0

Root Sum and Product:

• Sum of roots = –b/a
• Product of roots = c/a

Calculation:

Given: Equation is 2x² + kx + 8 = 0

Here, a = 2, b = k, c = 8

Discriminant D = k² – 4×2×8 = k² – 64

(a) Both roots are positive:

⇒ Sum of roots = –k/2 > 0 ⇒ k < 0

⇒ Product = 8/2 = 4 > 0 (Always positive)

⇒ For both roots positive

⇒ k < 0 and D > 0 ⇒ k² – 64 > 0 ⇒ k ∈ (–∞, –8) ∪ (8, ∞)

(b) Roots are real and equal:

⇒ D = 0

⇒ k² – 64 = 0

⇒ k = ±8

(c) Roots are imaginary:

⇒ D < 0

⇒ k² – 64 < 0

⇒ k² < 64

(d) Both roots are negative:

⇒ Sum of roots = –k/2 < 0 ⇒ k > 0

⇒ Product = 4 > 0 (Always positive)

⇒ For both roots negative

⇒ k > 0 and D = 0

⇒ k = 8 is a valid match when both roots equal and negative.

∴ Correct Matchings are: (a) → (iii) , (b) → (ii) , (c) → (iv) , ​(d) → (v)

Explanation:

Given:

(x - 1)2 + (x - 3)2 + (x - 5)2 = 0

⇒ x² – 2x + 1 + x² – 6x + 9 + x² – 10x + 25 = 0

⇒ 3x² – 18x + 35 = 0

Now, discriminant D =  (–18)² –4 × 3 × 35 

= – 96 < 0

Thus, The given equation does not have any real roots. 

∴ Option (a) is correct

Explanation:

Given:

tanα and tanβ are roots of 7x² – 6x + 1 = 0

Hence, 

⇒ tanα + tanβ = 6/7...(1)

⇒ tanα.tanβ = 1/7...(2)

⇒ tan(α+β) = \(\frac{tanα + tanβ}{1- tanα . tanβ} = \frac{\frac{6}{7}}{1- \frac{1}{7}} = 1\)

⇒ α +β = 45°

⇒ 2α +2β = 90°

It means the third angle of a triangle is 90°.

Hence, the triangle is right-angled.

Since

⇒tanα - tanβ = \(\sqrt{(tanα + tanβ)^2 - 4tanα .tanβ} = \frac{2\sqrt2}{7}\)

Also

⇒ tan(α -β) = \(\frac{tanα - tanβ}{1+ tanα .tanβ}\) = \(\frac{1}{2\sqrt2}\)

⇒ tan2(α -β) = \(\frac{2tan(α -β)}{1- tan^2(α-β)}\)

 = \(\frac{4\sqrt7}{7}\)

Thus 2α≠2β 

∴ Option (c) is correct.

Calculation:

Given,

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⇒ \(n^2 + pn + m = 0\) ..... (1)

Also, m is the root of equation of x2 + px + n = 0

⇒ \(m^2 +pm+ n =0\) ..... (2)

Now Eq. (i) – Eq. (ii), we get

⇒ \((n^2 – m^2) + p(n – m) – (n – m) = 0\)

⇒ \((n – m) [(n + m) + p – 1] = 0\)

⇒ \(n + m + p – 1 = 0\) ..... ( n ≠ m)

⇒ n + m + p = 1

∴ Option (c) is correct.

Concept:

For two distinct real solutions, D > 0,

Where D = b² - 4ac

Calculation:

⇒ k² - 4k > 0

⇒ k(k - 4) > 0

⇒ (k - 0) (k - 4) > 0

Case: I    k > 0 & k - 4 & 0

From here, we will get

k > 4

Case: II  k < 0 & k < 4

From here, we will get

k < 0

Hence, the solution will be k < 0 or k > 4.

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.