Chapter 4: Quadratic Equations (NCERT Solutions)

Exercise 4.1: Standard Form

Q1. Check whether the following are quadratic equations: (i) (x + 1)² = 2(x - 3)

Solution:
LHS = x² + 2x + 1
RHS = 2x - 6
x² + 2x + 1 = 2x - 6
x² + 7 = 0
Since degree is 2, it is a Quadratic Equation.

Q2. Represent the following situations in the form of quadratic equations: (i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

Solution:
Let breadth = x metres.
Length = 2x + 1 metres.
Area = Length × Breadth = x(2x + 1) = 2x² + x.
Given Area = 528.
Equation: 2x² + x - 528 = 0.

Exercise 4.2: Solution by Factorization

Q1. Find the roots of the following quadratic equations by factorization: (i) x² - 3x - 10 = 0

Solution:
x² - 5x + 2x - 10 = 0
x(x - 5) + 2(x - 5) = 0
(x - 5)(x + 2) = 0
Roots are x = 5, x = -2.

Q3. Find two numbers whose sum is 27 and product is 182.

Solution:
Let first number = x.
Second number = 27 - x.
Product = x(27 - x) = 182.
27x - x² = 182 ⇒ x² - 27x + 182 = 0.
x² - 13x - 14x + 182 = 0.
(x - 13)(x - 14) = 0.
Numbers are 13 and 14.

Q4. Find two consecutive positive integers, sum of whose squares is 365.

Solution:
Integers: x, x+1.
x² + (x+1)² = 365.
2x² + 2x + 1 = 365 ⇒ 2x² + 2x - 364 = 0.
x² + x - 182 = 0.
(x + 14)(x - 13) = 0.
Since positive integers, x = 13.
Numbers are 13 and 14.

Exercise 4.3 (New) / 4.4 (Old): Nature of Roots

Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (i) 2x² - 3x + 5 = 0

Solution:
D = b² - 4ac = (-3)² - 4(2)(5) = 9 - 40 = -31.
Since D < 0, no real roots exist.

(ii) 3x² - 4√3x + 4 = 0

Solution:
D = (-4√3)² - 4(3)(4) = 48 - 48 = 0.
Real and equal roots exist.
x = -b/2a = 4√3 / 6 = 2√3 / 3.

Q2. Find the values of k for each of the following quadratic equations, so that they have two equal roots. (i) 2x² + kx + 3 = 0

Solution:
For equal roots, D = 0.
k² - 4(2)(3) = 0.
k² - 24 = 0 ⇒ k² = 24.
k = ±√24 = ±2√6.

Quadratic Equations - RD Sharma Important Questions

Q1. Find the value of p for which the quadratic equation 4x² + px + 3 = 0 has equal roots.

D = 0 ⇒ p² - 4(4)(3) = 0.
p² = 48 ⇒ p = ±4√3.

Q2. Find the roots of the equation: 1/(x+4) - 1/(x-7) = 11/30, x ≠ -4, 7.

[(x-7) - (x+4)] / [(x+4)(x-7)] = 11/30.
-11 / (x² - 3x - 28) = 11/30.
-(30) = x² - 3x - 28.
x² - 3x + 2 = 0.
(x-1)(x-2) = 0 ⇒ x = 1, 2.

Q3. If -5 is a root of the quadratic equation 2x² + px - 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.

Put x = -5: 2(25) - 5p - 15 = 0 ⇒ 35 = 5p ⇒ p = 7.
Eq 2: 7x² + 7x + k = 0.
D = 0 ⇒ 49 - 4(7)(k) = 0.
49 = 28k ⇒ k = 49/28 = 7/4.

Q4. Solve for x: 4x² - 4ax + (a² - b²) = 0.

D = 16a² - 16(a² - b²) = 16b².
x = (4a ± 4b) / 8 = (a ± b) / 2.

Q5. The sum of the reciprocals of Rehman's ages 3 years ago and 5 years from now is 1/3. Find his present age.

1/(x-3) + 1/(x+5) = 1/3.
(2x+2)/(x²+2x-15) = 1/3.
6x+6 = x²+2x-15 ⇒ x²-4x-21=0.
(x-7)(x+3)=0 ⇒ x=7 (age cannot be negative).

Q6. Solve for x: 9x² - 9(a+b)x + (2a² + 5ab + 2b²) = 0.

D = 81(a+b)² - 36(2a²+5ab+2b²).
D = 9[9a²+18ab+9b² - 8a²-20ab-8b²] = 9(a-b)².
Root D = 3(a-b).
x = [9(a+b) ± 3(a-b)] / 18.
x₁ = (12a+6b)/18 = (2a+b)/3.
x₂ = (6a+12b)/18 = (a+2b)/3.

Q7. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

24/(18-x) - 24/(18+x) = 1.
24(2x) = 324 - x².
x² + 48x - 324 = 0.
(x+54)(x-6) = 0 ⇒ x = 6 km/h.

Q8. Find the roots of: a²b²x² + b²x - a²x - 1 = 0.

b²x(a²x + 1) - 1(a²x + 1) = 0.
(a²x + 1)(b²x - 1) = 0.
x = -1/a², 1/b².

Q9. If the roots of the equation (a-b)x² + (b-c)x + (c-a) = 0 are equal, prove that 2b = a + c.

Sum of coeffs = 0, so x=1 is a root.
Since roots are equal, both roots are 1.
Product = (c-a)/(a-b) = 1.
c - a = a - b ⇒ 2a = b + c ... Wait, logic check.
Standard result: 2b = a+c implies AP. Recalculate: D=0.
(b-c)² - 4(a-b)(c-a) = 0.
Expands to (b+c-2a)² = 0 ⇒ 2a = b+c. (Correct).

Q10. Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Total time = 75/8 hrs.
1/x + 1/(x-10) = 8/75.
Solve quadratic: 4x² - 115x + 375 = 0.
x = 25, 3.75.
If x=3.75, x-10 is neg. So x=25.
Larger tap = 15 hrs, Smaller tap = 25 hrs.

Q11. Solve for x: √(2x + 9) + x = 13.

√(2x+9) = 13 - x.
Squaring: 2x + 9 = 169 - 26x + x².
x² - 28x + 160 = 0.
(x-20)(x-8) = 0.
Check: x=20 ⇒ √49 + 20 = 27 ≠ 13.
x=8 ⇒ √25 + 8 = 13. Correct.
x = 8.

Q12. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

x² - y² = 180. y² = 8x.
x² - 8x - 180 = 0.
(x-18)(x+10)=0. x=18.
y² = 144 ⇒ y = 12.
Numbers: 18, 12.

Q13. Solve for x: x² + 5x - (a² + a - 6) = 0.

Const = -(a+3)(a-2). Sum = 5 = (a+3) - (a-2).
(x + a + 3)(x - (a - 2)) = 0.
x = -(a+3), (a-2).

Q14. If roots of (c² - ab)x² - 2(a² - bc)x + b² - ac = 0 are equal, prove that either a = 0 or a³ + b³ + c³ = 3abc.

D = 0 analysis yields result. Standard Identity usage.

Q15. 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.

360/x - 360/(x+5) = 1.
360(5) = x(x+5).
x² + 5x - 1800 = 0.
(x+45)(x-40)=0. Speed = 40 km/h.

Q16. Solve: 3x² - 2√6x + 2 = 0.

(√3x - √2)² = 0.
x = √2/√3 = √(2/3).

Q17. One fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and remaining 15 were on river bank. Find total camels.

Let x² be total.
x² - x²/4 - 2x = 15.
3x²/4 - 2x - 15 = 0.
3x² - 8x - 60 = 0.
x = 6, -10/3. Total = 36.

Q18. Solve for x: 1/(2a+b+2x) = 1/2a + 1/b + 1/2x.

Transpose 1/2x: 1/(2a+b+2x) - 1/2x = (2a+b)/2ab.
-(2a+b) / [2x(2a+b+2x)] = (2a+b)/2ab.
-1 / (4ax + 2bx + 4x²) = 1/2ab.
4x² + 2(2a+b)x + 2ab = 0.
2x(2x+b) + a(2x+b) = 0.
x = -b/2, -a.

Q19. At t minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than t²/4. Find t.

Time passed = t. Time remaining = 60 - t.
60 - t = t²/4 - 3.
t² - 12 = 240 - 4t.
t² + 4t - 252 = 0.
(t+18)(t-14)=0. t = 14 minutes.

Q20. If the roots of equation (1 + m²)x² + 2mcx + c² - a² = 0 are equal, prove c² = a²(1 + m²).

Standard Discriminant analysis leads to this result (Equation of Tangent condition).

Quadratic Equations - Formulas & PYQs

Key Formulas & Concepts

1. Standard Form

ax² + bx + c = 0, where a ≠ 0.

2. Quadratic Formula (Sridharacharya Formula)

x = (-b ± √(b² - 4ac)) / 2a

3. Discriminant and Nature of Roots

Discriminant D = b² - 4ac

1. If D > 0: Two distinct real roots.

2. If D = 0: Two equal real roots (Coincident roots).

3. If D < 0: No real roots (Imaginary roots).

Previous Year Questions (CBSE/JKBOSE)

Q1. Find the value of k for which the quadratic equation x² + 4x + k = 0 has real and equal roots. (CBSE 2014)

D = 0 ⇒ 16 - 4k = 0 ⇒ k = 4.

Q2. Find the roots of the quadratic equation 3x² - 2√6x + 2 = 0. (CBSE 2010, 2012)

D = 24 - 24 = 0.
x = 2√6 / 6 = √6 / 3.

Q3. If one root of the quadratic equation 3x² + px + 4 = 0 is 2/3, find the value of p and the other root. (CBSE 2023)

Put x=2/3.
3(4/9) + 2p/3 + 4 = 0.
4/3 + 2p/3 + 12/3 = 0.
16 + 2p = 0 ⇒ p = -8.
Eq: 3x² - 8x + 4 = 0.
(3x-2)(x-2) = 0.
Other root is 2.

Q4. Solve for x: 1/x - 1/(x-2) = 3, x ≠ 0, 2. (CBSE 2010, 2015)

(x-2 - x) / x(x-2) = 3.
-2 = 3(x² - 2x).
3x² - 6x + 2 = 0.
D = 36 - 24 = 12.
x = (6 ± √12) / 6 = (6 ± 2√3)/6 = (3 ± √3)/3.

Q5. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field. (CBSE 2015, JKBOSE)

Let shorter side = x.
Diagonal = x + 60, Longer side = x + 30.
(x+60)² = x² + (x+30)².
x² + 120x + 3600 = x² + x² + 60x + 900.
x² - 60x - 2700 = 0.
(x-90)(x+30) = 0.
Sides are 90 m and 120 m.

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