## Quadratic Equations - 10 Types of Question Patterns asked that You Must Know

**Q1**: Determine whether the given values are solutions of the given equation or not?

x² − 3x + 2 = 0, x = 2, x = −1

**Q2**: Find the value of k for which the given value is a solution of the given equation

x² − x(a + b) + k = 0, x = a

**Q3**: The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer.

**Q4**: Solve the give quadratic equation by factorization: (x - 4)(x + 2) = 0

**Q5**: Find the roots of the following quadratic equation (if they exist) by the method of completing

the square.

x² - 4√2 x + 6 = 0

**Q6**: Write the discriminant of the quadratic equation: 2x² - 5x + 3 = 0

**Q7**: Determine the nature of roots for the given quadratic equation:

x² + x + 2 = 0

**Q8**: Find the values of k for which the roots are real and equal for the given equation:

kx² + 4x + 1 = 0

**Q9**: (Word Problem) Find the consecutive numbers whose squares have the sum 85.

**Q10**: (Word Problem) A two-digit number is such that the products of its digits is 8. When 18 is subtracted from the number, the digits interchange their places. Find the number?

Solutions:

Answer 1:

LHS = x² − 3x + 2

At x = 2,

LHS = 2² - 3(2) + 2 = 4 - 6 + 2 = 6 - 6 = 0 = RHS

At x =-1,

LHS = (-1)² -3(-1) + 2 = 1 + 3 + 2 = 6 ≠ RHS

∴ x = 2 is the solution of the given quadratic equation.

Answer 2:

Given that x = a is a root of the given equation

x² − x(a + b) + k = 0

⇒ x = a satisfies the equation

⇒ (a)² − a(a + b) + k = 0

⇒ a² - a² - ab + k = 0

⇒ -ab + k = 0

⇒ k = ab (answer)

Answer 3:

Given that the smallest integer of 2 consecutive integer is denoted by x

⇒ The two integer will be x and (x + 1) and product is 306

⇒ x(x + 1) = 306

⇒ x² + x - 306 = 0 (answer)

Answer 4:

Given (x - 4)(x + 2) = 0

⇒ Either x - 4 = 0 or x + 2 = 0

⇒ x = 4 or x = -2

Thus, x = 4 and x = -2 are two roots of the equation (x - 4)(x + 2) = 0

Answer 5:

x² - 4√2 x + 6 = 0

⇒ x² - 2 × x × 2√2 + (2√2)² - (2√2)² + 6 = 0

⇒ (x - 2√2)² - (2√2)² + 6 = 0

⇒ (x - 2√2)² = (2√2)² - 6

⇒ (x - 2√2)² = 8 - 6

⇒ (x - 2√2)² = 2

⇒ x - 2√2 = ±√2

⇒ x - 2√2 = √2 OR x - 2√2 = -√2

⇒ x = 2√2 + √2 OR x = 2√2 -√2

⇒ x = 3√2 OR x = √2 (answer)

Answer 6:

2x² - 5x + 3 = 0

Here a = 2, b = -5 and c = 3

D = b² - 4ac

∴ D = (-5)² - 4(2)(3)

= 25 - 24

= 1 (answer)

Answer 7:

x² + x + 2 = 0

Here a = 1, b = 1, c = 2

D = b² - 4ac

∴ D = (1)² - 4(1)(2)

= 1 - 8

= -7

As D < 0, the equation has no real roots.

Answer 8:

*kx² + 4x + 1 = 0*

Here a = k, b = 4 and c = 1

It is given that, the equation has real and equal roots

D = b² - 4ac = 0

∴ (4)² - 4(k)(1) = 0

⇒ 16 - 4k = 0

⇒ 4k = 16

⇒ k = 4 (answer)

Answer 9:

Let the two consecutive natural numbers be ‘x’ and ‘x + 1’

⇒ Given that the sum of their squares is 85

ATQ,

x² + (x + 1)² = 85

⇒ x² + x² + 2x + 1 = 85

⇒ 2x² + 2x - 85 + 1 = 0

⇒ 2x² + 2x - 84 = 0

⇒ x² + x - 42 = 0

⇒ x² + 7x - 6x - 42 = 0 (By mid term splitting)

⇒ x(x + 7) -6(x + 7) = 0

⇒ (x + 7)(x - 6) = 0

⇒ x + 7 = 0 or x - 6 = 0

⇒ x = -7 or x = 6

Case I: x= -7

Ist number = -7

2nd number = -7 + 1 = -6

Case II: x = 6

Ist number = 6

2nd number = 6 + 1 = 7

The consecutive numbers are 6,7 and −6, −7. (answer)

Answer 10:

Let the two digits be x and x – 2

Given that the product of their digits is 8.

⇒ x(x – 2) = 8

⇒ x 2 − 2x − 8 = 0

⇒ x 2 − 4x + 2x − 8 = 0

⇒ x(x − 4) + 2(x − 4) = 0

⇒ (x − 4)(x + 2) = 0

⇒ x = 4 or x = −2

Considering the positive value x = 4, x – 2 = 2.

∴ The two digit number is 42.

**See also**: