Quadratic Equations - 1 Mark Problems
Important Formulas:
• Standard form: ax² + bx + c = 0
• Discriminant (D) = b² - 4ac
• Nature of roots:
- D > 0: Two distinct real roots
- D = 0: Two equal real roots
- D < 0: No real roots
• Sum of roots = -b/a
• Product of roots = c/a
• Discriminant (D) = b² - 4ac
• Nature of roots:
- D > 0: Two distinct real roots
- D = 0: Two equal real roots
- D < 0: No real roots
• Sum of roots = -b/a
• Product of roots = c/a
1
Check whether 1 and 3/2 are the roots of the equation 2x² – 5x + 3 = 0.
(J'15)
Step 1: Check x = 1
2(1)² - 5(1) + 3 = 2 - 5 + 3 = 0
Since the equation equals 0, x = 1 is a root.
Step 2: Check x = 3/2
2(3/2)² - 5(3/2) + 3 = 2(9/4) - 15/2 + 3
= 9/2 - 15/2 + 6/2 = (9 - 15 + 6)/2 = 0/2 = 0
Since the equation equals 0, x = 3/2 is also a root.
∴ Yes, both 1 and 3/2 are roots of the equation.
2
If b² – 4ac > 0 in ax² + bx + c = 0, (a ≠ 0); then what can you say about roots of the equation?
(M'16)
Step 1: Recall discriminant concept
For a quadratic equation ax² + bx + c = 0, the discriminant D = b² - 4ac determines the nature of roots.
∴ If b² - 4ac > 0, the equation has two distinct real roots.
3
Find the value of k, if 2 is one of the roots of the quadratic equation x² – kx + 6 = 0.
(J'16)
Step 1: Substitute x = 2 in the equation
(2)² - k(2) + 6 = 0
4 - 2k + 6 = 0
10 - 2k = 0
2k = 10
k = 5
∴ The value of k is 5.
4
Write the nature of roots of the quadratic equation 2x² – 5x + 6 = 0.
(M'17)
Step 1: Calculate discriminant
For 2x² - 5x + 6 = 0, a = 2, b = -5, c = 6
D = b² - 4ac = (-5)² - 4(2)(6) = 25 - 48 = -23
Step 2: Determine nature of roots
Since D < 0, the equation has no real roots.
∴ The equation has no real roots (complex roots).
5
Write the nature of the roots of the quadratic equation x² – 8x + 16 = 0.
(J'17)
Step 1: Calculate discriminant
For x² - 8x + 16 = 0, a = 1, b = -8, c = 16
D = b² - 4ac = (-8)² - 4(1)(16) = 64 - 64 = 0
Step 2: Determine nature of roots
Since D = 0, the equation has two equal real roots.
∴ The equation has two equal real roots.
6
Find sum and product of the roots of the quadratic equation x² – 4√3x + 9 = 0.
(M'18)
Step 1: Identify coefficients
For x² - 4√3x + 9 = 0, a = 1, b = -4√3, c = 9
Step 2: Calculate sum of roots
Sum of roots = -b/a = -(-4√3)/1 = 4√3
Step 3: Calculate product of roots
Product of roots = c/a = 9/1 = 9
∴ Sum of roots = 4√3, Product of roots = 9
7
Find the values of k for which the quadratic equation 4x² + 5kx + 25 = 0 has equal roots.
(J'18)
Step 1: Condition for equal roots
For equal roots, discriminant D = 0
Step 2: Calculate discriminant
For 4x² + 5kx + 25 = 0, a = 4, b = 5k, c = 25
D = b² - 4ac = (5k)² - 4(4)(25) = 25k² - 400
Step 3: Set D = 0 and solve for k
25k² - 400 = 0
25k² = 400
k² = 16
k = ±4
∴ The values of k are 4 and -4.
8
Find the roots of the quadratic equation x² + 2x – 3 = 0.
(M'19)
Step 1: Factor the equation
x² + 2x - 3 = 0
(x + 3)(x - 1) = 0
Step 2: Solve for x
x + 3 = 0 or x - 1 = 0
x = -3 or x = 1
∴ The roots are x = -3 and x = 1.
9
Find the discriminant of the quadratic equation 3x² – 5x + 2 = 0 and hence write the nature of its roots.
(J'19)
Step 1: Calculate discriminant
For 3x² - 5x + 2 = 0, a = 3, b = -5, c = 2
D = b² - 4ac = (-5)² - 4(3)(2) = 25 - 24 = 1
Step 2: Determine nature of roots
Since D > 0, the equation has two distinct real roots.
∴ Discriminant = 1, and the equation has two distinct real roots.
10
Is (x + 2)² = x² + 3 a Quadratic Equation? Justify.
(May 2022)
Step 1: Expand and simplify
(x + 2)² = x² + 3
x² + 4x + 4 = x² + 3
4x + 4 = 3
4x + 1 = 0
Step 2: Check if it's quadratic
A quadratic equation must have degree 2 (highest power of x is 2).
Here, the highest power of x is 1, so it's a linear equation.
∴ No, it's not a quadratic equation. It simplifies to a linear equation.
11
Is x(2x + 3) = x² + 5 a Quadratic Equation? Justify.
(Aug 2022)
Step 1: Expand and simplify
x(2x + 3) = x² + 5
2x² + 3x = x² + 5
2x² - x² + 3x - 5 = 0
x² + 3x - 5 = 0
Step 2: Check if it's quadratic
The simplified equation is x² + 3x - 5 = 0
This has degree 2 (highest power of x is 2), so it's a quadratic equation.
∴ Yes, it is a quadratic equation as it simplifies to x² + 3x - 5 = 0.
12
Solve the quadratic equation 2sin²θ – 3sinθ + 1 = 0 where 0° < θ ≤ 90°.
(Apr'23)
Step 1: Let y = sinθ
The equation becomes: 2y² - 3y + 1 = 0
Step 2: Factor the quadratic
2y² - 3y + 1 = 0
(2y - 1)(y - 1) = 0
Step 3: Solve for y
2y - 1 = 0 or y - 1 = 0
y = 1/2 or y = 1
Step 4: Solve for θ
sinθ = 1/2 or sinθ = 1
For sinθ = 1/2: θ = 30° (since 0° < θ ≤ 90°)
For sinθ = 1: θ = 90°
∴ The solutions are θ = 30° and θ = 90°.
Quadratic Equations - 2 Mark Problems
Important Formulas:
• Standard form: ax² + bx + c = 0
• Discriminant (D) = b² - 4ac
• For equal roots: D = 0
• Sum of roots = -b/a
• Product of roots = c/a
• Quadratic with roots α and β: x² - (α+β)x + αβ = 0
• Discriminant (D) = b² - 4ac
• For equal roots: D = 0
• Sum of roots = -b/a
• Product of roots = c/a
• Quadratic with roots α and β: x² - (α+β)x + αβ = 0
1
If 9x² + kx + 1 = 0 has equal roots, then find the value of k.
(M'16)
Step 1: Condition for equal roots
For equal roots, discriminant D = 0
Step 2: Identify coefficients
For 9x² + kx + 1 = 0, a = 9, b = k, c = 1
Step 3: Calculate discriminant
D = b² - 4ac = k² - 4(9)(1) = k² - 36
Step 4: Set D = 0 and solve for k
k² - 36 = 0
k² = 36
k = ±6
∴ The values of k are 6 and -6.
2
The sum of a number and its reciprocal is 10/3. Find the number.
(M'17)
Step 1: Formulate the equation
Let the number be x
Then, x + 1/x = 10/3
Step 2: Multiply through by 3x to eliminate denominators
3x(x + 1/x) = 3x(10/3)
3x² + 3 = 10x
3x² - 10x + 3 = 0
Step 3: Solve the quadratic equation
3x² - 10x + 3 = 0
3x² - 9x - x + 3 = 0
3x(x - 3) - 1(x - 3) = 0
(3x - 1)(x - 3) = 0
x = 1/3 or x = 3
∴ The number is 3 or 1/3.
3
Is it possible to design a rectangular Garden, whose length is twice of its breadth and area is 200 m²? If so, find its length and breadth.
(J'17)
Step 1: Set up variables
Let breadth = x meters
Then length = 2x meters
Step 2: Formulate the equation
Area = length × breadth = 2x × x = 2x²
Given area = 200 m²
So, 2x² = 200
Step 3: Solve for x
2x² = 200
x² = 100
x = 10 (taking positive value as length can't be negative)
Step 4: Find dimensions
Breadth = x = 10 m
Length = 2x = 20 m
∴ Yes, it is possible. The garden has length 20 m and breadth 10 m.
4
If the equation kx² – 2kx + 6 = 0 has equal roots, then find the value of k.
(M'18)
Step 1: Condition for equal roots
For equal roots, discriminant D = 0
Step 2: Identify coefficients
For kx² - 2kx + 6 = 0, a = k, b = -2k, c = 6
Step 3: Calculate discriminant
D = b² - 4ac = (-2k)² - 4(k)(6) = 4k² - 24k
Step 4: Set D = 0 and solve for k
4k² - 24k = 0
4k(k - 6) = 0
k = 0 or k = 6
Step 5: Check validity
If k = 0, the equation becomes 6 = 0, which is not quadratic
So we discard k = 0
∴ The value of k is 6.
5
Without calculating the roots of x² - 5x + 6 = 0, explain the nature of roots.
(J'18)
Step 1: Identify coefficients
For x² - 5x + 6 = 0, a = 1, b = -5, c = 6
Step 2: Calculate discriminant
D = b² - 4ac = (-5)² - 4(1)(6) = 25 - 24 = 1
Step 3: Analyze discriminant
Since D > 0 and is a perfect square, the roots are:
1. Real and distinct
2. Rational numbers
∴ The equation has two distinct rational roots.
6
Write the Quadratic equation, whose roots are 2 + √3 and 2 - √3.
(M'19)
Step 1: Find sum of roots
Sum = (2 + √3) + (2 - √3) = 4
Step 2: Find product of roots
Product = (2 + √3)(2 - √3) = 4 - 3 = 1
(Using identity: (a+b)(a-b) = a² - b²)
Step 3: Form the quadratic equation
Quadratic equation with roots α and β is: x² - (α+β)x + αβ = 0
So, x² - (4)x + (1) = 0
x² - 4x + 1 = 0
∴ The required quadratic equation is x² - 4x + 1 = 0.
7
Find the roots of quadratic equation x² + 4x + 3 = 0 by "completing square method".
(J'19)
Completing Square Method:
Step 1: Write the equation
x² + 4x + 3 = 0
Step 2: Move constant term to RHS
x² + 4x = -3
Step 3: Add square of half the coefficient of x to both sides
Coefficient of x = 4, half of it = 2, square = 4
x² + 4x + 4 = -3 + 4
x² + 4x + 4 = 1
Step 4: Write LHS as perfect square
(x + 2)² = 1
Step 5: Take square root on both sides
x + 2 = ±1
Step 6: Solve for x
x + 2 = 1 or x + 2 = -1
x = -1 or x = -3
∴ The roots are x = -1 and x = -3.
8
Shashanka said that (x + 1)² = 2(x – 3) is a quadratic equation. Do you agree?
(J'19)
Step 1: Expand and simplify
(x + 1)² = 2(x - 3)
x² + 2x + 1 = 2x - 6
Step 2: Bring all terms to one side
x² + 2x + 1 - 2x + 6 = 0
x² + 7 = 0
Step 3: Check if it's quadratic
A quadratic equation must have degree 2 (highest power of x is 2).
Here, the equation is x² + 7 = 0, which has degree 2.
∴ Yes, I agree with Shashanka. It is a quadratic equation.
9
Write a Quadratic Equation, whose roots are 2 – √3 and 2 + √3.
(Aug 22)
Step 1: Find sum of roots
Sum = (2 - √3) + (2 + √3) = 4
Step 2: Find product of roots
Product = (2 - √3)(2 + √3) = 4 - 3 = 1
(Using identity: (a+b)(a-b) = a² - b²)
Step 3: Form the quadratic equation
Quadratic equation with roots α and β is: x² - (α+β)x + αβ = 0
So, x² - (4)x + (1) = 0
x² - 4x + 1 = 0
∴ The required quadratic equation is x² - 4x + 1 = 0.
Quadratic Equations - 4 Mark Problems
Important Formulas:
• Standard form: ax² + bx + c = 0
• Discriminant (D) = b² - 4ac
• For equal roots: D = 0
• Sum of roots = -b/a
• Product of roots = c/a
• Quadratic with roots α and β: x² - (α+β)x + αβ = 0
• Area of square = side²
• Perimeter of square = 4 × side
• Distance = Speed × Time
• Discriminant (D) = b² - 4ac
• For equal roots: D = 0
• Sum of roots = -b/a
• Product of roots = c/a
• Quadratic with roots α and β: x² - (α+β)x + αβ = 0
• Area of square = side²
• Perimeter of square = 4 × side
• Distance = Speed × Time
1
If the sum of the areas of two squares is 468 m² and the difference of their perimeters is 24 m, then find the measurements of their sides.
(J'15)
Step 1: Set up variables
Let side of first square = x meters
Let side of second square = y meters
Step 2: Form equations from given conditions
Area of first square = x²
Area of second square = y²
Sum of areas: x² + y² = 468 ...(1)
Perimeter of first square = 4x
Perimeter of second square = 4y
Difference of perimeters: 4x - 4y = 24 ...(2)
Step 3: Simplify equation (2)
4x - 4y = 24
4(x - y) = 24
x - y = 6 ...(3)
Step 4: Solve the system of equations
From (3): x = y + 6
Substitute in (1): (y + 6)² + y² = 468
y² + 12y + 36 + y² = 468
2y² + 12y + 36 - 468 = 0
2y² + 12y - 432 = 0
Divide by 2: y² + 6y - 216 = 0
Step 5: Solve the quadratic equation
y² + 6y - 216 = 0
y² + 18y - 12y - 216 = 0
y(y + 18) - 12(y + 18) = 0
(y + 18)(y - 12) = 0
y = -18 or y = 12
Since side cannot be negative, y = 12
Step 6: Find x
x = y + 6 = 12 + 6 = 18
∴ The sides of the squares are 18 m and 12 m.
2
Sum of the squares of two consecutive positive even integers is 100; find those numbers by using quadratic equations.
(M'16)
Step 1: Set up variables
Let first even integer = x
Then next consecutive even integer = x + 2
Step 2: Form the equation
Sum of squares: x² + (x + 2)² = 100
Step 3: Expand and simplify
x² + (x² + 4x + 4) = 100
2x² + 4x + 4 = 100
2x² + 4x + 4 - 100 = 0
2x² + 4x - 96 = 0
Divide by 2: x² + 2x - 48 = 0
Step 4: Solve the quadratic equation
x² + 2x - 48 = 0
x² + 8x - 6x - 48 = 0
x(x + 8) - 6(x + 8) = 0
(x + 8)(x - 6) = 0
x = -8 or x = 6
Step 5: Check for positive integers
Since we need positive integers, x = 6
Then x + 2 = 8
∴ The two consecutive positive even integers are 6 and 8.
3
If -4 is a common root for the quadratic equations 2x² + px + 8 = 0 and p(x² + x) + k = 0, find the value of k.
(J'17)
Step 1: Substitute x = -4 in first equation
2x² + px + 8 = 0
2(-4)² + p(-4) + 8 = 0
2(16) - 4p + 8 = 0
32 - 4p + 8 = 0
40 - 4p = 0
4p = 40
p = 10
Step 2: Expand second equation
p(x² + x) + k = 0
px² + px + k = 0
Step 3: Substitute x = -4 and p = 10 in second equation
10(-4)² + 10(-4) + k = 0
10(16) - 40 + k = 0
160 - 40 + k = 0
120 + k = 0
k = -120
∴ The value of k is -120.
4
Sum of squares of two consecutive even numbers is 580. Find the numbers by writing a suitable quadratic equation.
(M'18)
Step 1: Set up variables
Let first even number = x
Then next consecutive even number = x + 2
Step 2: Form the equation
Sum of squares: x² + (x + 2)² = 580
Step 3: Expand and simplify
x² + (x² + 4x + 4) = 580
2x² + 4x + 4 = 580
2x² + 4x + 4 - 580 = 0
2x² + 4x - 576 = 0
Divide by 2: x² + 2x - 288 = 0
Step 4: Solve the quadratic equation
x² + 2x - 288 = 0
x² + 18x - 16x - 288 = 0
x(x + 18) - 16(x + 18) = 0
(x + 18)(x - 16) = 0
x = -18 or x = 16
Step 5: Find the numbers
If x = 16, then x + 2 = 18
If x = -18, then x + 2 = -16
∴ The numbers are 16 and 18, or -18 and -16.
5
If a number when increased by 12, equals 160 times of its reciprocal, then find the numbers.
(J'18)
Step 1: Set up variables
Let the number be x
Its reciprocal is 1/x
Step 2: Form the equation
Number increased by 12: x + 12
160 times its reciprocal: 160 × (1/x) = 160/x
Equation: x + 12 = 160/x
Step 3: Multiply both sides by x
x(x + 12) = 160
x² + 12x = 160
x² + 12x - 160 = 0
Step 4: Solve the quadratic equation
x² + 12x - 160 = 0
x² + 20x - 8x - 160 = 0
x(x + 20) - 8(x + 20) = 0
(x + 20)(x - 8) = 0
x = -20 or x = 8
∴ The numbers are 8 and -20.
6
Sum of the areas of two squares is 850 m². If the difference of their perimeters is 40 m. Find the sides of the two squares.
(M'19)
Step 1: Set up variables
Let side of first square = x meters
Let side of second square = y meters
Step 2: Form equations from given conditions
Area of first square = x²
Area of second square = y²
Sum of areas: x² + y² = 850 ...(1)
Perimeter of first square = 4x
Perimeter of second square = 4y
Difference of perimeters: 4x - 4y = 40 ...(2)
Step 3: Simplify equation (2)
4x - 4y = 40
4(x - y) = 40
x - y = 10 ...(3)
Step 4: Solve the system of equations
From (3): x = y + 10
Substitute in (1): (y + 10)² + y² = 850
y² + 20y + 100 + y² = 850
2y² + 20y + 100 - 850 = 0
2y² + 20y - 750 = 0
Divide by 2: y² + 10y - 375 = 0
Step 5: Solve the quadratic equation
y² + 10y - 375 = 0
y² + 25y - 15y - 375 = 0
y(y + 25) - 15(y + 25) = 0
(y + 25)(y - 15) = 0
y = -25 or y = 15
Since side cannot be negative, y = 15
Step 6: Find x
x = y + 10 = 15 + 10 = 25
∴ The sides of the squares are 25 m and 15 m.
7
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
(J'19)
Step 1: Set up variables
Let actual speed = x km/h
Increased speed = (x + 5) km/h
Distance = 360 km
Step 2: Calculate times
Actual time = Distance/Speed = 360/x hours
Time at increased speed = 360/(x + 5) hours
Step 3: Form the equation
Actual time - New time = 1 hour
360/x - 360/(x + 5) = 1
Step 4: Multiply both sides by x(x + 5)
360(x + 5) - 360x = x(x + 5)
360x + 1800 - 360x = x² + 5x
1800 = x² + 5x
x² + 5x - 1800 = 0
Step 5: Solve the quadratic equation
x² + 5x - 1800 = 0
x² + 45x - 40x - 1800 = 0
x(x + 45) - 40(x + 45) = 0
(x + 45)(x - 40) = 0
x = -45 or x = 40
Since speed cannot be negative, x = 40
∴ The speed of the train is 40 km/h.
8
The numerator of a fraction is 3 less than its denominator. If 2 is added to both numerator and denominator, the sum of the new fraction formed and original fraction is 29/20 then find the original fraction.
(Jun'23)
Step 1: Set up variables
Let denominator = x
Then numerator = x - 3
Original fraction = (x - 3)/x
Step 2: Form the new fraction
When 2 is added to both numerator and denominator:
New numerator = (x - 3) + 2 = x - 1
New denominator = x + 2
New fraction = (x - 1)/(x + 2)
Step 3: Form the equation
Sum of original and new fraction = 29/20
(x - 3)/x + (x - 1)/(x + 2) = 29/20
Step 4: Find common denominator and simplify
[(x - 3)(x + 2) + x(x - 1)] / [x(x + 2)] = 29/20
[(x² - x - 6) + (x² - x)] / [x(x + 2)] = 29/20
[2x² - 2x - 6] / [x(x + 2)] = 29/20
Divide numerator and denominator by 2:
[x² - x - 3] / [x(x + 2)] = 29/20
Step 5: Cross multiply
20(x² - x - 3) = 29x(x + 2)
20x² - 20x - 60 = 29x² + 58x
20x² - 29x² - 20x - 58x - 60 = 0
-9x² - 78x - 60 = 0
Multiply by -1: 9x² + 78x + 60 = 0
Divide by 3: 3x² + 26x + 20 = 0
Step 6: Solve the quadratic equation
3x² + 26x + 20 = 0
Using quadratic formula: x = [-b ± √(b² - 4ac)] / (2a)
a = 3, b = 26, c = 20
Discriminant = 26² - 4(3)(20) = 676 - 240 = 436
√436 = √(4 × 109) = 2√109
x = [-26 ± 2√109] / 6 = [-13 ± √109] / 3
Step 7: Find the fraction
x = [-13 + √109] / 3 (taking positive value as denominator can't be negative)
Numerator = x - 3 = [-13 + √109] / 3 - 3 = [-13 + √109 - 9] / 3 = [-22 + √109] / 3
Original fraction = ([-22 + √109] / 3) / ([-13 + √109] / 3) = (-22 + √109) / (-13 + √109)
∴ The original fraction is (-22 + √109) / (-13 + √109).
