Chapter 4: Quadratic Equations

Exercise 4.1

1. Check whether the following are quadratic equations:

(i) (x + 1)² = 2(x – 3)

(ii) x² – 2x = (–2)(3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

(iv) (x – 3)(2x + 1) = x(x + 5)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) x² + 3x + 1 = (x – 2)²

(vii) (x + 2)³ = 2x(x² – 1)

(viii) x³ – 4x² – x + 1 = (x – 2)³

(i) Simplifies to: x² + 7 = 0 — Quadratic

(ii) Simplifies to: x² – 4x + 6 = 0 — Quadratic

(iii) Simplifies to: –3x + 1 = 0 — Linear, not quadratic

(iv) Simplifies to: x² – 10x – 3 = 0 — Quadratic

(v) Simplifies to: x² – 11x + 8 = 0 — Quadratic

(vi) Simplifies to: 7x – 3 = 0 — Linear, not quadratic

(vii) Simplifies to: –x³ + 6x² + 14x + 8 = 0 — Cubic, not quadratic

(viii) Simplifies to: 2x² – 13x + 9 = 0 — Quadratic

2. Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is 528 m². The length (L) is one more than twice the breadth (B). Find the quadratic equation.

(ii) The product of two consecutive positive integers is 306. Find the quadratic equation.

(iii) Rohan’s mother is 26 years older than him. The product of their ages 3 years from now will be 360. Find the quadratic equation.

(iv) A train travels 480 km at uniform speed. If speed was 8 km/h less, it would take 3 hours more. Find the quadratic equation.

(i) Let breadth = x meters, length = 2x + 1 meters.

Area = length × breadth → x(2x + 1) = 528

2x² + x – 528 = 0

This is the quadratic equation.

(ii) Let the first integer = x, second integer = x + 1.

Product = x(x + 1) = 306

x² + x – 306 = 0

This is the quadratic equation.

(iii) Let Rohan’s age = x years.

Mother’s age = x + 26 years.

Three years later:

(x + 3)(x + 26 + 3) = 360

(x + 3)(x + 29) = 360

Expanding: x² + 32x + 87 = 360

x² + 32x + 87 – 360 = 0 → x² + 32x – 273 = 0

This is the quadratic equation.

(iv) Let speed = x km/h.

Time taken = distance / speed = 480 / x hours.

If speed reduced by 8 km/h, time taken = 480 / (x – 8) hours.

According to problem:

480/(x – 8) – 480/x = 3

Multiply through by x(x – 8):

480x – 480(x – 8) = 3x(x – 8)

480x – 480x + 3840 = 3x² – 24x

3840 = 3x² – 24x

3x² – 24x – 3840 = 0

Divide entire equation by 3:

x² – 8x – 1280 = 0

This is the quadratic equation.

Exercise 4.2

1. Find the roots of the following quadratic equations by factorisation:

(i) x² – 3x – 10 = 0

(ii) 2x² + x – 6 = 0

(iii) 2x² + 7x + 5 = 0

(iv) 2x² – x + 1/8 = 0

(v) 100x² – 20x + 1 = 0

(i) x² – 3x – 10 = 0

Factors of -10 with sum -3: -5 and 2

(x - 5)(x + 2) = 0 → Roots: x = 5, x = -2

(ii) 2x² + x – 6 = 0

Multiply a*c = 2 * (-6) = -12

Find factors of -12 that add to b=1: 4 and -3

Rewrite: 2x² + 4x - 3x - 6 = 0

Group: 2x(x + 2) - 3(x + 2) = 0

(2x - 3)(x + 2) = 0 → Roots: x = 3/2, x = -2

(iii) 2x² + 7x + 5 = 0

Multiply a*c = 2*5=10

Factors of 10 that add to 7: 5 and 2

Rewrite: 2x² + 5x + 2x + 5 = 0

Group: x(2x + 5) + 1(2x + 5) = 0

(x + 1)(2x + 5) = 0 → Roots: x = -1, x = -5/2

(iv) 2x² – x + 1/8 = 0

Multiply entire equation by 8 to clear fraction:

16x² – 8x + 1 = 0

Try to factor or use quadratic formula (factorisation is tough here)

Roots by formula: x = [8 ± √(64 – 64)] / 32 = 8/32 = 1/4 (double root)

(v) 100x² – 20x + 1 = 0

Try factorisation:

Rewrite as (10x – 1)² = 0 → Roots: x = 1/10 (double root)

2. Solve the problems given in Example 1.

(Refer to your textbook for Example 1 solutions.)

3. Find two numbers whose sum is 27 and product is 182.

Let numbers be x and y.

x + y = 27

xy = 182

Form quadratic: t² – (sum)t + product = 0

t² – 27t + 182 = 0

Factor: 182 factors that add to 27 → 13 and 14

(t – 13)(t – 14) = 0 → Numbers: 13, 14

4. Find two consecutive positive integers, sum of whose squares is 365.

Let first integer be x, second = x + 1.

x² + (x + 1)² = 365

x² + x² + 2x + 1 = 365

2x² + 2x + 1 = 365

2x² + 2x – 364 = 0

x² + x – 182 = 0

Factor: (x + 14)(x – 13) = 0

Positive integer: x = 13, next integer = 14

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Let base = x cm, altitude = x – 7 cm.

Hypotenuse = 13 cm.

By Pythagoras: x² + (x – 7)² = 13²

x² + x² – 14x + 49 = 169

2x² – 14x + 49 = 169

2x² – 14x – 120 = 0

Divide by 2: x² – 7x – 60 = 0

Factor: (x – 12)(x + 5) = 0

Positive side: x = 12 cm

Altitude = 12 – 7 = 5 cm

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.

Let the number of articles produced = x.

Cost per article = 2x + 3 rupees.

Total cost = number × cost per article = x(2x + 3) = 90

2x² + 3x – 90 = 0

Factorise:

Multiply a*c = 2 * (-90) = -180

Factors of -180 adding to 3: 15 and -12

Rewrite: 2x² + 15x – 12x – 90 = 0

Group: x(2x + 15) – 6(2x + 15) = 0

(2x + 15)(x – 6) = 0

Possible x: –15/2 (ignore, negative) or 6

Number of articles = 6

Cost per article = 2(6) + 3 = 12 + 3 = ₹ 15

Exercise 4.3

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x² – 3x + 5 = 0

(ii) 3x² – 4√3 x + 4 = 0

(iii) 2x² – 6x + 3 = 0

For quadratic ax² + bx + c = 0, discriminant Δ = b² – 4ac.

(i) 2x² – 3x + 5 = 0

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

Δ = (-3)² – 4×2×5 = 9 – 40 = -31 < 0

Roots are complex (no real roots).

(ii) 3x² – 4√3 x + 4 = 0

a=3, b=–4√3, c=4

Δ = (–4√3)² – 4×3×4 = 16×3 – 48 = 48 – 48 = 0

Roots are real and equal.

Root = –b/2a = 4√3 / (2×3) = 2√3/3

(iii) 2x² – 6x + 3 = 0

a=2, b=–6, c=3

Δ = (–6)² – 4×2×3 = 36 – 24 = 12 > 0

Roots are real and unequal.

Roots = [6 ± √12]/(2×2) = [6 ± 2√3]/4 = (3 ± √3)/2

2. Find the values of k for each of the following quadratic equations, so that they have two equal roots:

(i) 2x² + kx + 3 = 0

(ii) kx(x – 2) + 6 = 0

For equal roots, discriminant Δ = 0.

(i) 2x² + kx + 3 = 0

a=2, b=k, c=3

Δ = k² – 4×2×3 = k² – 24 = 0

k² = 24 → k = ±2√6

(ii) kx(x – 2) + 6 = 0

Rewrite: kx² – 2kx + 6 = 0

a = k, b = –2k, c = 6

Δ = (–2k)² – 4×k×6 = 4k² – 24k = 0

4k² – 24k = 0 → 4k(k – 6) = 0

k = 0 or k = 6

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.

Let breadth = x m, length = 2x m.

Area = length × breadth = 2x × x = 2x² = 800

2x² = 800 → x² = 400 → x = 20 m (breadth)

Length = 2×20 = 40 m

Yes, it is possible.

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Let their present ages be x and y.

x + y = 20 → y = 20 – x

Four years ago, ages: x – 4 and y – 4

Product: (x – 4)(y – 4) = 48

Substitute y:

(x – 4)(20 – x – 4) = 48

(x – 4)(16 – x) = 48

Expand: 16x – x² – 64 + 4x = 48

–x² + 20x – 64 = 48

–x² + 20x – 112 = 0

Multiply by –1: x² – 20x + 112 = 0

Discriminant Δ = 400 – 448 = –48 < 0

No real roots → Situation not possible.

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.

Let length = l, breadth = b.

Perimeter P = 2(l + b) = 80 → l + b = 40 → l = 40 – b

Area A = l × b = 400

Substitute l:

b(40 – b) = 400

40b – b² = 400

Rearranged: b² – 40b + 400 = 0

Discriminant Δ = 1600 – 1600 = 0

Roots are equal → b = 40/2 = 20 m

Length = 40 – 20 = 20 m

Yes, possible. The park is a square 20 m × 20 m.