Quadratic Equation: Practice Questions | CSAT Preparation - UPSC PDF Download

Given: 
(p – x)(p – y) = 1
Also, (x - y)=√5
Formula Used:
(x – y)³ = x³ – y³ – 3xy(x – y)
Calculation:
(p – x)(p – y) = 1

⇒ x - y = √5      ----(2)
Add and subtract 'p' in left side of eq. (2)
⇒ -p + x + p – y = √5 
⇒ (p − y) − (p − x) = √5      ----(3)
Put the value of (p - x) from eq. (1) to eq. (3)

Taking cube both sides, we have –

Given:

6x² + 3x² – 5x + 1

Calculation:

6x² + 3x² – 5x + 1

⇒ 9x² – 5x + 1

Let 'a' and 'b' be two roots of the equations

As we know,

Sum of roots (α + β) = (-b)/a = 5/9

Product of roots (αβ) = c/a = 1/9

According to the question

⇒ 1/α + 1/β

⇒ (α + β)/αβ

⇒ [5/9] / [1/9] = 5

Concept:

Roots of a quadratic equation satisfy the equation, so by putting the value of one root,

One can find the unknown variable and hence the other root.

Calculation:

Let us put x = 3 in the equation x² – 12x + k = 0,

⇒ 9 – 36 + k = 0

⇒ k = 27

Putting the value of k in the equation,

we get: x² – 12x + 27 = 0

⇒ x² – 9x – 3x + 27 = 0

⇒ x(x – 9) – 3 (x – 9) = 0

⇒ (x – 3)(x – 9) = 0

⇒ x = 3 and 9

∴ Other root of the equation is 9

Given:

x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 = 0     ----(1)

Calculation:

Multiplying by x, we have –

⇒ x⁷ +x⁶ + x⁵ + x⁴ + x³ + x² + x = 0      ----(2)

Equation (2) – (1), we have –

⇒ x⁷ – 1 = 0

⇒ x⁷ = 1

According to question –

The value of x⁴² + x⁸⁴ = 

Given:

One root of the equation is 5 = 2√5

Concept: 

If one root of the quadratic equation is in this form (a + √b) then the other roots must be conjugate (a - √b) and vice-versa.

Quadratic equation: x² - (sum of root) + (product of root) = 0

Calculation: 

Quadratic equation = x² - (α + β)x + α β = 0

Now, Quadratic equation = x² - 10x + 5 = 0 

Hence, required quadratic equation is x² - 10x + 5 = 0

Given:

The given equation is ax² + x + b = 0

Concept used:

General form of the quadratic equation is ax² + x + b = 0

Condition for roots,

For equal and real roots, b² – 4ac = 0 

For unequal and real roots, b² – 4ac > 0

For imaginary roots, b² – 4ac < 0  

Calculation:

For equal and real roots, b² – 4ac = 0 

⇒ b² = 4ac

After comparing with the general form of the quadratic equation we'll get

b = 1, a = a and c = b

Then, b² = 4ac

⇒ 1 = 4ab

⇒ ab = 1/4

∴ The correct relation is ab = 1/4

Given:

5x² + 2x + Q = 2

Given α = 1/β ⇒ α.β = 1 ----(i)

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 is given by:

α + β = − b/a = −(coefficient of x/coefficient of x²)

The product of the roots is given by:

α × β = c/a = (constant term /coefficient of x²)

Calculation:

Let the roots of 5x² + 2x + Q - 2 = 0 are α and β

According to the question,

α = 1/β 

⇒  α.β = 1 

Compare with general equation ax² + bx + c = 0

a = 5, b = 2, c = Q - 2

⇒  (Q – 2)/5 = 1

⇒ Q - 2 = 5

⇒ Q = 7

Hence, Q² = 7² = 49.

Given:

3x² – ax + 6 = ax² + 2x + 2

⇒ 3x² – ax² – ax – 2x + 6 – 2 = 0

⇒ (3 – a)x² – (a + 2)x + 4 = 0

Concept Used:

If a quadratic equation (ax2 + bx + c=0) has equal roots, then discriminant should be zero i.e. b2 – 4ac = 0

Calculation:

⇒ D = B² – 4AC = 0

⇒ (a + 2)² – 4(3 – a)4 = 0

⇒ a² + 4a + 4 – 48 + 16a = 0

⇒ a² + 20a – 44 = 0

⇒ a² + 22a – 2a – 44 = 0

⇒ a(a + 22) – 2(a + 22) = 0

⇒ a = 2, -22

∴ Positive integral solution of a = 2

Given:

Two roots are 2 + √5 and 2 - √5.

Concept used:

The quadratic equation is:

x² - (Sum of roots)x + Product of roots = 0

Calculation:

Let the roots of the equation be A and B.

A = 2 + √5 and B = 2 - √5

⇒ A + B = 2 + √5 + 2 - √5 = 4

⇒ A × B = (2 + √5)(2 - √5) = 4 - 5 = -1

Then equation is

∴ x² - 4x - 1 = 0

Mistake Points:

For a quadratic equation, ax² + bx + c = 0,

Sum of the roots = (-b/a) = 4/1

Product of the roots = c/a = -1/1

Then, b = -4

So, the sign of coefficient of x is negative. 

Given:

x² – x – 1 = 0

Formula used:

If the given equation is ax² + bx + c = 0

Then Sum of roots = -b/a

And Product of roots = c/a

Calculation:

As α and β are roots of x² – x – 1 = 0, then

⇒ α + β = -(-1) = 1

⇒ αβ = -1

Now, if (α/β) and (β/α) are roots then,

⇒ Sum of roots = (α/β) + (β/α)

⇒ Sum of roots = (α² + β²)/αβ

⇒ Sum of roots = [(α + β)² – 2αβ]/αβ

⇒ Sum of roots = (1)² – 2(-1)]/(-1) = -3

⇒ Product of roots = (α/β) × (β/α) = 1

Now, then the equation is,

⇒ x² – (Sum of roots)x + Product of roots = 0

⇒ x² – (-3)x + (1) = 0

⇒ x² + 3x + 1 = 0

For I:

x² – 2x – 80 = 0

⇒ x² – 10x + 8x – 80 = 0

⇒ x(x – 10) + 8(x – 10) = 0

⇒ (x + 8) (x – 10) = 0

⇒ x = -8, 10

For II:

y² – 10y – 171 = 0

⇒ y² – 19y + 9y – 171 = 0

⇒ y(y – 19) + 9(y – 19) = 0

⇒ (y – 19) (y + 9) = 0

⇒ y = 19, -9

Comparison between x and y (via Tabulation):

So x = y or the relationship cannot be established.

For I:

x² + 8x – 48 = 0

⇒ x² + 12x – 4x – 48 = 0

⇒ x(x + 12) – 4(x + 12) = 0

⇒ (x + 12) – 4(x + 12) = 0

⇒ x = -12, 4

For II:

y² – 8y – 105 = 0

⇒ y² – 15y + 7y – 105 = 0

⇒ y(y – 15) + 7(y – 15) = 0

⇒ (y – 15)(y + 7) = 0

⇒ y = 15, -7

Comparison between x and y (via Tabulation):

So x = y or relation between x and y can not be established.

For I:

x² + x – 90 = 0

⇒ x² + 10x – 9x – 90 = 0

⇒ x(x + 10) – 9(x + 10) = 0

⇒ (x – 9) (x + 10) = 0

⇒ x = 9, -10

For II:

y² + 4y – 96 = 0

⇒ y² + 12y – 8y – 96 = 0

⇒ y(y + 12) – 8(y + 12) = 0

⇒ (y + 12) (y – 8) = 0

⇒ y = -12, 8

Comparison between x and y (via Tabulation):

So x = y or relation between x and y can not be established.

Concept:

If roots of ax² + bx + c = 0 are equal then 

D = b² - 4ac = 0

Calculation:

Given the quadratic equation ax² + 2bx + c = 0,

the roots are equal if the discriminant is zero:

(2b)² - 4ac = 0

⇒ 4b² = 4ac

⇒ b² = ac

⇒ b = √ac

So, the correct answer is (2) a, b, c are in Geometric Progression (GP).

Given:

x = -2/3

Calculation:

9x² – 3x – 11 = ?

⇒ 9 × (-2/3)² – 3(-2/3) – 11 = ?

⇒ 9 × (4/9) – (-2) – 11 = ?

⇒ 4 + 2 – 11 = ?

⇒ 6 – 11 = ?

⇒ -5 = ?

∴ The required value is -5