Test: Quadratic Equations: Common Roots (June 9) - JEE MCQ

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

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:

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

Concept:

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

then sum of roots = α + β = -b/a

and product of roots = αβ = c/a

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: 

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 Standard Form of a Quadratic Equation is ax2 + bx + c = 0

Sum of roots = −b/a

Product of roots = c/a

Formulae

α2 + β2 = (α + β)2 - 2α.β

Calculation:

If α and β are the roots of the equation ax2 + bx + b = 0.

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:

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.

⇒ m = -5

Thus, m = -5 and n = 30.

Concept:

If two roots of the equation ax² + bx + c = 0 are equal, then 

b² - 4ac = 0

Calculation:

If the equation (m - n)x² + (n - l)x + l - m = 0 has equal roots, 

⇒ (n - l)² - 4(m - n)(l - m) = 0

⇒ n² + l² - 2ln - 4lm + 4m² + 4ln - 4mn = 0

⇒ n² + l2 + 4m² + 2ln - 4lm - 4mn = 0

⇒ (n + l - 2m)² = 0

⇒ n + l = 2m

∴ The correct answer is option (2).

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:

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

Let α be a common root of both the 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 = 18/q-p

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