Sum and Product of Roots MCQ Quiz - Objective Question with Answer for Sum and Product of Roots - Download Free PDF

Given:

α and β  are the zeroes of a quadratic polynomial x2 - 25x + 150 = 0

Concept:

If ax2 + bx + c = 0 be a quadratic equation, then 

Sum of roots = -b/a

Product of roots = c/a

Calculation:

Here, 

x2 - 25x + 150 

a = 1, b = -25, c = 150

Sum of roots  (α + β)= \(\frac{-(-25)}{1}\) = 25

Product of roots αβ  = \(\frac{150}{1}\) = 150

\(\frac{1}{α}+\frac{1}{β}\) = \(\frac{(α+β)}{αβ}\) = \(\frac{25}{150}\) = \(\frac{1}{6}\)

∴ Option 4 is correct

Calculation:

In ∠ABC,

\(\angle \mathrm{A}+\frac{\pi}{2}+\angle \mathrm{B}=180^{\circ}\)

∴ ∠A + ∠B + ∠C = 180°

∴ ∠A + ∠B = \(\frac{\pi}{2}\)

∴ \(\frac{\angle \mathrm{A}}{2}+\frac{\angle \mathrm{B}}{2}=\frac{\pi}{4}\)

Now, tan \(\left(\frac{\mathrm{A}}{2}\right)\) and tan \(\left(\frac{\mathrm{B}}{2}\right)\) are roots of equation

a₁x² + b₁x + c₁ = 0 ...[Given]

∴ Sum of roots = \(\frac{-b_1}{a_1}\)

\(\tan \left(\frac{\mathrm{A}}{2}\right)+\tan \left(\frac{\mathrm{B}}{2}\right)=\frac{-\mathrm{b}_1}{\mathrm{a}_1}\)

Also, tan \(\left(\frac{\mathrm{A}}{2}\right)\) tan \(\left(\frac{\mathrm{B}}{2}\right)\) = \(\frac{c_1}{\mathrm{a}_1}\)

Using \(\tan \left(\frac{A}{2}+\frac{B}{2}\right)=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}\), we get

\(\tan \left(\frac{\pi}{4}\right)=\frac{\frac{-b_1}{a_1}}{1-\frac{s_1}{a_1}}\)

\(1=\frac{-b_1}{a_1-c_1}\)

a₁ - c₁ = -b₁

a₁ + b₁ = c₁ 

∴ The require relation is a1 + b1 = c1.

The correct answer is Option 1.

Calculation

a, b is solutions of x2 + px + 1 = 0

⇒ a + b = -p, ab = 1.... (i)

c, d is solution of x2 + qx + 1 = 0

⇒ c + d = - q, cd = 1

Now (a - c)(b - c) and (a + d)(b + d) are

the roots of x² + ax + β = 0

(a − c)(b  - c)(a + d)(b + d) = β 

⇒ (ab - ac - bc + c²)(ab + ad + bd + d²) = β 

⇒ {1 - c(a + b) + c²)(1 + d(a + b) + d²) = β 

⇒ (1 + pc + c²)(1 - pd + d²) = β 

⇒ 1 - pd + d² + pc - p²cd + pcd² + c² - pc²d + c²d² = β

⇒ 1 - pd + d² + pc - p² + pd + c² - pc + 1 = [∵ cd = 1] 

⇒ 2 + d² + c² - p² = β 

⇒ 2cd + c² + d² - p² = β 

⇒ (c + d)² - p² = β 

⇒ q² - p² = β[∵ (c + d) = -q]

Hence option 4 is correct

Concept Used:

For a cubic equation of the form \(Ax^3 + Bx^2 + Cx + D = 0\) with roots \(\alpha, \beta, \gamma\):

1. Sum of roots: \(\alpha + \beta + \gamma = -\frac{B}{A}\)

2. Sum of product of roots taken two at a time: \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{C}{A}\)

3. Product of roots: \(\alpha\beta\gamma = -\frac{D}{A}\)

Calculation:

For the given equation \(x^3 + ax^2 + bx + c = 0\):

⇒ \(\alpha + \beta + \gamma = -a\)

⇒ \(\alpha\beta + \beta\gamma + \gamma\alpha = b\)

⇒ \(\alpha\beta\gamma = -c\)

 \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \)

⇒ \(\frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma}\)

⇒ \(\frac{b}{-c}\)

∴ \(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = -\frac{b}{c}\)

Hence option 2 is correct

Concept Used:

Vieta's formulas for cubic equations.

\(\sum \alpha = \alpha + \beta + \gamma = -\frac{b}{a}\)

\(\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\)

\(\alpha\beta\gamma = -\frac{d}{a}\)

Calculation

Given:

Equation: \(2x^3 - 3x^2 + 5x - 7 = 0\)

Here, a = 2, b = -3, c = 5, d = -7

\(\sum \alpha = \alpha + \beta + \gamma = -\frac{-3}{2} = \frac{3}{2}\)

\(\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = \frac{5}{2}\)

\(\alpha\beta\gamma = -\frac{-7}{2} = \frac{7}{2}\)

We need to find \(\sum \alpha^2\beta^2 = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2\)

\((\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 + 2(\alpha\beta^2\gamma + \alpha^2\beta\gamma + \alpha\beta\gamma^2)\)

⇒ \((\sum \alpha\beta)^2 = \sum \alpha^2\beta^2 + 2\alpha\beta\gamma(\beta + \alpha + \gamma)\)

⇒ \((\sum \alpha\beta)^2 = \sum \alpha^2\beta^2 + 2\alpha\beta\gamma(\sum \alpha)\)

⇒ \(\sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha)\)

⇒ \(\sum \alpha^2\beta^2 = (\frac{5}{2})^2 - 2(\frac{7}{2})(\frac{3}{2})\)

⇒ \(\sum \alpha^2\beta^2 = \frac{25}{4} - \frac{42}{4}\)

⇒ \(\sum \alpha^2\beta^2 = \frac{25-42}{4}\)

⇒ \(\sum \alpha^2\beta^2 = \frac{-17}{4}\) 

Hence option 1 is correct

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: 

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:

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

∴ \(\rm \dfrac{\pi}{2}\)

Concept:

For a quadratic equation ax2 + bx + c = 0

The sum of the roots = \(\rm-{b\over a}\)

The product of the roots = \(\rm{c\over a}\)

Calculation:

Let the other root be β  

Given equation is x(x + 1) + 1 = 0

⇒ x² + x + 1 = 0

a = 1, b = 1 and c = 1

As k is the root of the equation

⇒ k2 + k + 1 = 0

⇒ k2 = -1 - k     .....(i)

The sum of the roots = \(-{1\over1}\) = -1

⇒ β + k = -1

⇒ β = -1 - k      .....(ii)

From equation (i) and (ii), we get

⇒ β = k²

∴ The other root = k²

Given equation is x(x + 1) + 1 = 0

Factor of (x2 + x + 1) = 0

\({\rm{x}} = {\rm{\;}}\frac{{ - 1{\rm{\;}} \pm {\rm{\;}}\sqrt {{1^2} - 4{\rm{\;}} \times 1{\rm{\;}} \times 1} }}{{2{\rm{\;}} \times 1}} = {\rm{\;}}\frac{{ - 1{\rm{\;}} \pm {\rm{i}}\sqrt 3 }}{2}\)

\(⇒ {\rm{x}} = {\rm{\;}}\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}{\rm{\;or\;\;}}\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}\)

⇒ x = ω or ω2

Consider k = ω 

∴ The other root = ω² = k2

Concept:

General Quadratic Equation

ax2 + bx + c = 0

• Product of roots (αβ) = c/a
• Sum of roots (α + β) = - b/a
• Formula to find Roots,  \(\displaystyle \rm x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
• sin 3θ = 3 sinθ - 4 sin³θ 

Calculation:

Given: 4x2 + 2x - 1 = 0

a = 4, b = 2, c = -1

\(\displaystyle x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

⇒ \(\displaystyle x = {-2 \pm \sqrt{2^2-4(4)(-1)} \over 2(4)}\)

⇒ \(\displaystyleα = {-1 + \sqrt{5} \over 4}\) and \(\displaystyle β = {-1 - \sqrt{5} \over 4}\)

As we know that \(\displaystyle sin\ 18° = {-1 + \sqrt{5} \over 4}\) and \(\displaystyle sin\ 54° = {1 + \sqrt{5} \over 4}\)

So, We can say that α = sin18° and β = - sin54°      ------(i)

Now, using the formula, sin 3θ = 3 sinθ - 4 sin3θ 

On putting θ = 18° in the above trigonometric formula, we get 

⇒ sin 54° = 3 sin18° - 4 sin318°      ------(ii)

From (i) and (ii), we get 

⇒ - β = 3α - 4α³

⇒ β = 4α3 - 3α

∴ The correct relation is β = 4α3 - 3α.

Concept:

The roots of a quadratic equation ax2 + bx + c = 0 is given by:

\(\rm x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Calculation:

Given quadratic equation:

\(\rm 2x^2-\sqrt5x-2 = 0\)

a = 2, b = \(-\sqrt5\) and c = -2

∴ \(\rm x = {-(-\sqrt5)\pm \sqrt{(-\sqrt5)^2- 4\times 2\times(-2)} \over 2\times2}\)

\(\rm x = {\sqrt5\pm \sqrt{5+16} \over 4}\)

\(\rm x = {\sqrt5+\sqrt{21} \over 4},{\sqrt5-\sqrt{21} \over 4}\)

∴ The roots of the equation are real and one positive and other negative

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: 4x2 + 8x – β = 0 and roots are -5α and 3

Now, sum of roots:

⇒ -5α + 3 = -(8)/4 = -2 

⇒ -5α = -5

⇒ α = 1

Now, Product of the roots:

⇒ (-5α)(3) = -β/4

⇒ - 15 α  = -β/4

⇒ 15 α = β/4

⇒ 15 = β/4 (∵ α = 1)

∴  β = 60

Hence, option (2) is correct. 

Concept:

For a quadratic equation ax² + bx + c = 0

Sum of roots = -b/a

and the product of roots = c/a

Calculation:

If the sum of roots = product of roots

⇒ \(\frac{-b}{a}=\frac ca\)

⇒ -b = c

So, there can be infinitely many quadratic equations with -b = c and a ≠ 0

∴ The correct option is (4).

Concept:

If α and β are the roots of equation , ax2 + bx + c =0 

Sum of roots (α + β) = \(\rm \frac{-b}{a}\)  

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

(x + y)2 = x2 + y2 + 2xy .

Calculation:

Given: f (x) = x2 - 5x + 6

Comparing f(x) with ax2 + bx + c =0 , we have , a = 1 , b= -5 and c=  6. 

Now, sum of roots =  α + β = \(\rm \frac{-b}{a}\) = \(\rm \frac{-(-5)}{1}\) = 5

And product of roots αβ = \(\rm \frac{c}{a}\) = \(\rm \frac{6}{1}\) = 6 . 

Now, α2β + β2α = αβ ( α+ β ) 

= 6 × 5 

= 30

The correct option is 2. 

Concept: 

Product of roots:

Let α and β are roots of ax2 + bx + c = 0,then α × β = c/a

Calculation: 

Here, x2 + x + 2 = 0 comparing with  ax2 + bx + c = 0

We get, a = 1, b = 1, c =2

Product of roots = α × β = c/a = 2

\(\frac{{{α ^{10}} + {β ^{10}}}}{{{α ^{ - 10}} + {β ^{ - 10}}}}\)

\(\begin{array}{l} ⇒ \frac{\alpha^{10}+\beta^{10}}{\frac{1}{\alpha^{10}}+\frac{1}{\beta^{10}}} \\ ⇒ \frac{\alpha^{10}+\beta^{10}}{\frac{\beta^{10}+\alpha^{10}}{\alpha^{10} \cdot \beta^{10}}} \\ ⇒ \alpha^{10} \cdot \beta^{10} \\ ⇒ (\alpha \cdot \beta)^{10} \\ ⇒ (2)^{10} \end{array}\)

⇒ 1024

Hence, option (3) is correct.