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

Calculation:

Given,

The equation is $x^2-x+1=0$

We need to find the value of the following expression:

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

The equation $x^2-x+1=0$ is solved as follows:

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

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:

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

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

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

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.

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

$\frac{1}{\alpha }+\frac{1}{\beta }$ = $\frac{(\alpha +\beta )}{\alpha \beta }$ = $\frac{25}{150}$ = $\frac{1}{6}$

∴ Option 4 is correct

Calculation:

In ∠ABC,

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

∴ ∠A + ∠B + ∠C = 180°

∴ ∠A + ∠B = $\frac{\pi }{2}$

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

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 $\tan (\frac{A}{2}+\frac{B}{2})=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}\tan \frac{B}{2}}$, 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 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: $\alpha +\beta =-\frac{\mathrm{b}}{\mathrm{a}}=-\frac{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{f}\mathrm{i}\mathrm{c}\mathrm{i}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{f}\mathrm{x}}{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{f}\mathrm{i}\mathrm{c}\mathrm{i}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{f}{\mathrm{x}}^2}$ 

The product of the roots is given by:  $\alpha \beta =\frac{\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}}{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{f}\mathrm{i}\mathrm{c}\mathrm{i}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{f}{\mathrm{x}}^2}$

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.${\displaystyle \frac{\pi }{4}}$

∴ ${\displaystyle \frac{\pi }{2}}$

Concept:

For a quadratic equation ax2 + bx + c = 0

The sum of the roots = $-\frac{\mathrm{b}}{\mathrm{a}}$

The product of the roots = $\frac{\mathrm{c}}{\mathrm{a}}$

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 = $-\frac{1}{1}$ = -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

$\mathrm{x}=\frac{-1\pm \sqrt{1^2-4\times 1\times 1}}{2\times 1}=\frac{-1\pm \mathrm{i}\sqrt{3}}{2}$

$\Rightarrow \mathrm{x}=\frac{-1+\mathrm{i}\sqrt{3}}{2}\mathrm{o}\mathrm{r}\frac{-1-\mathrm{i}\sqrt{3}}{2}$

⇒ 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,  ${\displaystyle \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{a}\mathrm{c}}}{2\mathrm{a}}}$
• sin 3θ = 3 sinθ - 4 sin³θ 

Calculation:

Given: 4x2 + 2x - 1 = 0

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

${\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

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

⇒ ${\displaystyle \alpha =\frac{-1+\sqrt{5}}{4}}$ and ${\displaystyle \beta =\frac{-1-\sqrt{5}}{4}}$

As we know that ${\displaystyle sin18°=\frac{-1+\sqrt{5}}{4}}$ and ${\displaystyle sin54°=\frac{1+\sqrt{5}}{4}}$

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:

$\mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{a}\mathrm{c}}}{2\mathrm{a}}$

Calculation:

Given quadratic equation:

$2{\mathrm{x}}^2-\sqrt{5}\mathrm{x}-2=0$

a = 2, b = $-\sqrt{5}$ and c = -2

∴ $\mathrm{x}=\frac{-(-\sqrt{5})\pm \sqrt{(-\sqrt{5}{)}^2-4\times 2\times (-2)}}{2\times 2}$

$\mathrm{x}=\frac{\sqrt{5}\pm \sqrt{5+16}}{4}$

$\mathrm{x}=\frac{\sqrt{5}+\sqrt{21}}{4},\frac{\sqrt{5}-\sqrt{21}}{4}$

∴ 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

⇒ $\frac{-b}{a}=\frac{c}{a}$

⇒ -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) = $\frac{-\mathrm{b}}{\mathrm{a}}$  

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

(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 = $\frac{-\mathrm{b}}{\mathrm{a}}$ = $\frac{-(-5)}{1}$ = 5

And product of roots αβ = $\frac{\mathrm{c}}{\mathrm{a}}$ = $\frac{6}{1}$ = 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

$\frac{{\alpha }^{10}+{\beta }^{10}}{{\alpha }^{-10}+{\beta }^{-10}}$

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

⇒ 1024

Hence, option (3) is correct.