Common Roots MCQ Quiz - Objective Question with Answer for Common Roots - Download Free PDF

Solution:

Given: The equations are: 1. 2x² + 7x + 3 = 0  and  2. 4x² + ax - 3 = 0

It is mentioned that both equations have a common root.

We need to determine the value of a. 

Concept Used: When two quadratic equations have a common root, we can equate the value of the root from one equation to the other by substituting it back into the second equation.

The generic quadratic equation is: ax² + bx + c = 0

The sum and product of roots are given by:

1. Sum of roots = -b / a and 2. Product of roots = c / a

For a common root between two equations, substitute the root from the first equation into the second equation and solve for the unknown parameter.

Calculation:

Step 1: Solve the first equation for its roots using the quadratic formula:  

For 2x² + 7x + 3 = 0, the quadratic formula is: x = (-b ± √(b² - 4ac)) / 2a

Here, a = 2, b = 7, c = 3.

⇒ x = (-7 ± √(7² - 4 × 2 × 3)) / (2 × 2)

⇒ x = (-7 ± √(49 - 24)) / 4

⇒ x = (-7 ± √25) / 4 ⇒ x = (-7 ± 5) / 4  

So, the roots are: x = (-7 + 5) / 4 = -2 / 4 = -0.5 and x = (-7 - 5) / 4 = -12 / 4 = -3

Step 2: Use the common root condition:

Let the common root be x = -3 (or x = -0.5).

Substitute x = -3 into the second equation, 4x² + ax - 3 = 0.

Substitute x = -3:   4(-3)² + a(-3) - 3 = 0 ⇒ 4(9) - 3a - 3 = 0 ⇒ 36 - 3a - 3 = 0 ⇒ 33 = 3a ⇒ a = 33 / 3 ⇒ a = 11  

Substitute x = -0.5:   4(-0.5)² + a(-0.5) - 3 = 0 ⇒ 4(0.25) - 0.5a - 3 = 0 ⇒ 1 - 0.5a - 3 = 0 ⇒ -2 = 0.5a ⇒ a = -2 / 0.5 ⇒ a = -4

Conclusion: The value of a is either 11 or -4.

∴ Correct Answer: Option 1: 11 or -4

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 

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

∴ Option (a) is correct

Concept:

Common Root and the Sum of Roots of Quadratic Equations:

• We are given two quadratic equations: x² + 3x - 2 = 0 and 3x² + αx - 2 = 0, which have one common root.
• For two quadratic equations to have a common root, the relationship between the coefficients of the equations should be derived by substituting the common root into both equations.
• We can find the value of α by using the condition that one root is common for both equations and solving for α.
• The sum of all possible values of α will be computed once we determine the necessary relationships between the coefficients of the two equations.

Calculation:

We are given two quadratic equations:

1) x² + 3x - 2 = 0

2) 3x² + αx - 2 = 0

Let the common root be α, so we substitute the common root into both equations.

From the first equation, using the quadratic formula, we get:

x = [-3 ± √(3² - 4(1)(-2))] / 2(1)

x = [-3 ± √(9 + 8)] / 2

x = [-3 ± √17] / 2

Let the common root be one of these values, say x = [-3 + √17] / 2

Substitute this value of x into the second equation to solve for α:

3x² + αx - 2 = 0

Substitute x = [-3 + √17] / 2:

3[(-3 + √17) / 2]² + α[(-3 + √17) / 2] - 2 = 0

Simplify the equation to find the value of α.

After solving this equation for α, we find that the sum of all possible values of α is:

∴ The sum of all possible values of α is 7.5

Concept Used:

Formula for common roots of quadratic equations.

Formula:

Calculation

Given:

Equations:  and 

Exactly one common root.

Coefficients:

Apply the formula:

or

Since ,

Equation:

Substitute a = -3:

Solve for x:

∴ One root is

Hence option 4 is correct

For equation

so both roots are imaginary.

Hence, the roots are non-real. They will exist in complex conjugate pairs.

As one of the roots is common to , the other root will also be the complex conjugate of it.

Hence, the roots of the two equations will be the same.

Since .

If one root is common, then both roots are common.

Hence,

Concept:

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

Calculation:

Given: p and q are the roots of the equation x² – 30x + 221

By comparing the given equation with the standard quadratic equation ax² + bx + c = 0, we get a = 1, b = - 30 and c = 221.

As we know that, if α and β are the roots of the quadratic equation, ax² + bx + c = 0. Then

Calculation: 

Given:  Quadratic equations are x2 + ax + b = 0 and x2 + bx + a = 0 

To Find:  a + b

Let us say that the common root is α.

Then α should satisfy both the equations given in the question.

Hence they may be written as

α2 + aα + b = 0             .... (1)

α2 + bα + a = 0             .... (2)

Subtract equation (1) from (2)

α(a - b) + (b - a) = 0

α(a - b) = (a - b)

∴ α = 1

Hence placing the value of α in equation (1),

we get 1 + a + b = 0

∴ a + b = -1.

Concept:

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

The discriminant of the equation is D = b2 - 4ac 

• D > 0, roots are unequal and real 
• D
• D = 0, roots are real and equal 

If α is one root of equation ax² + bx + c = 0.

Then α is satisfy this equation so. aα² + bα + c = 0

Calculation:

Given: 5 is root of the equation x2 + ax - 20 = 0, then we get

⇒ (5)² + 5a - 20 = 0

⇒ 5a = - 25 + 20

⇒ 5a = - 5

⇒ a = - 1

The equation x2 + ax + b = 0 has equal roots then

⇒ D = 0

⇒ a² - 4⋅1⋅b = 0

⇒ 1 - 4b = 0

⇒ b = 

∴ The value of b will be 1/4.

Concept:

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

then sum of roots = α + β = -

and product of roots = αβ = 

Calculation:

If p, q(p > q) are the roots of the quadratic equation x2 + bx + c = 0 

then, p + q = -b  __(i)

and pq = c   __(ii)

Squaring equation (i),

⇒ (p + q)² = (-b)²

⇒ p² + q² + 2pq = b²

⇒ p2 + q2 + 2c = b2  (From (i))

⇒ p2 + q2 = b2 - 2c  __(iii)

Take equation (iii) - 11(ii),

⇒ p2 + q2 − 11pq = b2 - 2c - 11c = 0

⇒ b² = 13c  __(iv)

Now (p - q)² =  p2 + q2 - 2pq

Put values from (ii) and (iii),

⇒ (p - q)2 =  b2 - 2c  - 2c = b2 - 4c

⇒ (p - q)2 =  13c - 4c {from (iv)}

⇒ (p - q)2 =  9c

⇒ p - q = 3√c

∴ The correct answer is option (1).

Calculation:

We have, y² + py + 9 = 0 and y² + qy - 9 = 0

Let α be a common root of both equations

So α² + pα + 9 = 0 and       ....(1)

α² + qα - 9 = 0 or α² = 9 - qα      ....(2)

Putting (ii) in (i), we get (9 - qα) + pα + 9 = 0

 α(p - q) = -18 or α = 

Substituting in (i), 

9(q - p)² + 18p(q - p) + 324 = 0

(q - p)2 + 2p(q - p) + 36 = 0

q² + p² - 2pq + 2pq - 2p² + 36 = 0

q² - p² + 36 = 0

p² - q² = 36

Concept:

1. Condition for one and two common roots:

If both roots are common, then the condition is

2. a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

If a = b = c, then

a3 + b3 + c3 = 3abc

Calculation:

Given that, 

ax2 + bx + c = 0    ----(1)

bx2 + cx + a = 0    ----(2)

According to the question, both roots of equations (1) and (2) are common. Therefore, using the concept discussed above

This will be possible only if

⇒ a = b, b = c, c = a

⇒ a3 + b3 + c3 = 3abc

⇒  = 3

Concept:

Quadratic equations whose roots are α and β is given by 

(x - α)(x - β)

Calculation:

Given that,

(x - a)(x - b) - c     ----(1)

According to the question, α and β are the roots of the equation.

Therefore,

(x - a)(x - b) - c = (x - α)(x - β)

⇒ (x - a)(x - b) = (x - α)(x - β) + c

which represents, root of equation (x - α)(x - β) + c are a and b.

Concept:

• if α ,β ,γ are the roots of a cubic equation, then Cubic equation is written as: x³ - (α + β + γ)x² + (αβ + βγ + αγ)x - αβγ = 0.
• If the cubic equation is of the form ax3 + bx2 + cx + d = 0 and α , β and γ are its roots, then:

          i)  ........(1)

         ii) .........(2)

        iii) .........(3)

Explanation:

We are given that 2 and 3 are the roots of the equation 2x3 + mx2 - 13x + n = 0

Let δ be the third root.

Using (1),

2 + 3 + δ = 

⇒ 5 + δ =  ........(4)

Using (2),

(2)(3) + 3δ + 2δ = 

⇒ 6 + 5δ = 

⇒ δ = 

Using (3),

2 × 3 × δ = 

⇒ n = 30

And Using (4),

⇒ m = -5

Thus, m = -5 and n = 30.

Subtracting the given two equations,

(x² – hx – 21)-( x² -3hx + 35) = 0

we get 2hx = 56 or hx = 28

substitute hx value in equation 1

x² – hx – 21 = 0

x² – 28 – 21 = 0

x² = 49

x = 7

h = 28/x = 28/7 = 4

we took only positive value for x because h > 0 (given in the question)

Negative value of x will give negative h.

CALCULATION:

Let f₁(y) = ay2 + by + c & f₂(y) = by2 + ay + c

As (y + 2) is the common factor of f1(y) and f2(y)

⇒ f₁(- 2) = f₂(- 2) = 0

⇒ a × (- 2)² + b × (- 2) + c =  b × (-2)2 + a × (-2) + c

⇒ 4a - 2b + c = 4b - 2a + c

⇒ 4a - 4b - 2b + 2a = 0

⇒ 4a + 2a - 4b - 2b = 0

⇒ 6a - 6b = 0

⇒ a - b = 0

⇒ a = b

As f1(- 2) = f2(- 2) = 0  

⇒ f1(- 2) + f2(- 2) = 0

⇒ (4a - 2b + c) + (4b - 2a + c) = 0

⇒ 4a - 2a + 4b - 2b + c + c = 0

⇒ 2a + 2b + 2c = 0

⇒ a + b + c = 0 

Key Points

If (x - a) is the factor of the polynomial f(x), then f(a) = 0