Test: Quadratic Equations: Relation between Roots (June 7) - JEE MCQ

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

⇒ 1024
Hence, option (3) is correct.

Concept:

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

Sum of roots (α + β) = −b/a  

Product of roots  (αβ) = 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 = −ba = −(−5)1 = 5

And product of roots αβ = ca = 61 = 6 . 

Now, α²β + β²α = αβ ( α+ β ) 

= 6 × 5 

= 30

The correct option is 2. 

Concept:

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

Calculation:

Given quadratic equation:

2x² - √5x - 2 = 0

a = 2, b = −√5 and c = -2

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

Concept:

For a quadratic equation ax² + bx + c = 0

The sum of the roots = −b/a

The product of the roots = c/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 = −1/1 = -1

⇒ β + k = -1

⇒ β = -1 - k      .....(ii)

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

⇒ β = k²

∴ The other root = k²

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

Given: 

If α and β are zeroes of the polynomial x² + 3x + 4

Concept Used: 

A quadratic equation ax² + bx + c = 0, if α and β are the roots :

α + β = - b/a and α × β = c/a

Formula Used:

α² + β² = (α + β)² - 2αβ

Calculation: 

a = 1, b = 3 & c = 4

⇒ α + β = - 3

⇒ α × β = 4

Substituting the given values

⇒ α² + β² = (-3)² - 2 × 4

⇒ α² + β² = 9 - 8

⇒ α² + β² = 1

∴ Correct answer is 1.

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. 

Given:

f(x) = x² + 5x + k, and α and β are its zeros,

The relation is  α² + β² - αβ = 22. 

Concept Used:

The zeros of a quadratic polynomial fx = ax² + bx + c, where a ≠ 0

The sum of the roots α + β = -b/a 

The product of the roots α × β = c/a 

Formula Used:

α² + β² = (α + β)² - 2αβ)

Calculation:

a = 1, b = 5 & c = k

α + β = - 5 and α × β = k

The given This can be re-written by using the identities from above and the formula

⇒ {(α + β)² - 2αβ} - αβ = 22

⇒ (-5)² - 3 × k = 22,

⇒ 25 - 3k = 22,

⇒ 3k = 25 - 22

⇒ 3k = 3 

⇒ k = 1

∴ Correct answer is k = 1

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, 
• sin 3θ = 3 sinθ - 4 sin³θ 

Calculation:

Given: 4x² + 2x - 1 = 0

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

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 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.π/4

∴ π/2