Quadratic Equations Class 10 Exe- 5A RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-5. Step by step solutions of questions related root of equation and method of factorization with zero product rule as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10.
Quadratic Equations Class 10 Exe- 5A RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-5
| Board | ICSE |
| Subject | Maths |
| Class | 10th |
| Chapter-5 | Quadratic Equations |
| Writer/ Book | RS Aggarwal |
| Topics | Solution of Exe-5A Questions |
| Academic Session | 2024-2025 |
Solution of Exe-5A Questions
Quadratic Equations Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-5
Page- 52,53
Exercise- 5A
( Root of Quadratic Equations and method of factorization with zero product rule )
Que-1: Find which of the following are the solution of the equation 6x²-x-2 =0 ?
(i) 1/2 (ii) -1/2 (iii) 2/3
Solution- 6x²−x−2 = 0
Here, a = 6, b = −1, c = 2
x = [−b±√b²−4ac]/2a
= [−(−1)±√(−1)²−4(6)(−2)]/2(6)
= [1±√1+48]/12
= [1±√49]/12
= [1±7]/12
x = (1+7)/12 and x = (1−7)/12
∴ x = 2/3 and x = −1/2
So, 2/3 and -1/2 are the solution of 6x²−x−2 = 0
Que-2: Determine whether x = -1/3 and x = 2/3 are the solution of the equation 9x²-3x-2 = 0.
Solution- Given equation is 9x2 – 3x – 2 = 0; and x = -1/3, or 𝑥 = 2/3
Substitute x = -1/3 in the L.H.S.
L.H.S. = 9(-1/3)²-3×(-1/3)-2
= 9×1/9+1-2
= 2 – 2
= 0
= R.H.S.
Hence, x = -1/3 is a solution of the equation.
Again now put x = 2/3
L.H.S. = 9(2/3)²-3(2/3)-2
= 9×4/9-2-2
= 4 – 4
= 0
= R.H.S.
Hence, x = 2/3 is a solution of the equation.
Solve the following equations by factorization :
Que-3: 16x² = 25
Solution- 16x² = 25
x² = 25/16
x = √(25/16)
x = ±5/4
x = 5/4 or -5/4 Ans.
Que-4: x²+2x = 24
Solution- x²+2x = 24
x²+2x-24 = 0
x²+6x-4x+24 = 0
x(x+6)-4(x+6) = 0
(x-4)(x+6) = 0
x-4 = 0 and x+6 = 0
x = 4,-6 Ans.
Que-5: x²-x = 156
Solution- x²-x = 156
x²-x-156 = 0
x²+12x-13x-156 = 0
x(x+12)-13(x+12) = 0
(x+12)(x-13) = 0
x+12 = 0 and x-13 = 0
x = -12,13 Ans.
Que-6: x²-11x = 42
Solution- x²-11x = 42
x²-11x-42 = 0
x²+3x-14x-42 = 0
x(x+3)-14(x+3) = 0
(x+3)(x-14) = 0
x+3 = 0 and x-14 = 0
x = -3,14 Ans.
Que-7: x²-7x+10 = 0
Solution- x²-7x+10 = 0
x²-2x-5x+10 = 0
x(x-2)-5(x-2) = 0
(x-2)(x-5) = 0
x-2 =0 and x-5 = 0
x = 2,5 Ans.
Que-8: x²+18x = 40
Solution- x²+18x = 40
x²+8x-40 = 0
x²+20x-2x-40 = 0
x(x+20)-2(x+20)
(x-2)(x+20) = 0
x-2 = 0 and x+20 = 0
x = 2,-20 Ans.
Que-9: x²+17 = 18x
Solution- x²+17 = 18x
x²-18x+17 = 0
x²-17x-x+17 = 0
x(x-17)-1(x-17) = 0
(x-17)(x-1) = 0
x-17 = 0 and x-1 = 0
x = 17,1 Ans.
Que-10: 3x² = 5x
Solution- 3x² = 5x
3x²-5x = 0
x(3x-5) = 0
x = 0 and (3x-5) = 0
x = 0, 5/3 Ans.
Que-11: (x+3)(x-3) = 27
Solution- (x+3)(x-3) = 27
x²-3x+3x-9 = 27
x² = 27+9
x² = 36
x = √36
x = ±6
x = -6,6 Ans.
Que-12: x²-30x+216 = 0
Solution- x²-30x+216 = 0
x²-18x-12x+216 = 0
x(x-18)-12(x-18) = 0
(x-12)(x-18) = 0
(x-12) = 0 and (x-18) = 0
x = 12,18 Ans.
Que-13: 12x²+29x+14 = 0
Solution- 12x²+29x+14 = 0
12x²+8x+21x+14 = 0
4x(3x+2)+7(3x+2) = 0
(4x+7)(3x+2) = 0
(4x+7) = 0 and (3x+2) = 0
x = -7/4, -2/3 Ans.
Que-14: 2x²-7x = 39
Solution- 2x²-7x = 39
2x²-7x-39 = 0
2x²+6x-13x-39 = 0
2x(x+3)-13(x+3) = 0
(2x-13)(x+13) = 0
(2x-13) = 0 and (x+13) = 0
x = 13/2, -3 Ans.
Que-15: 10x² = 9x+7
Solution- 10x² = 9x+7
10x²-9x-7 = 0
10x²+5x-14x-7 = 0
5x(2x+1)-7(2x+1) = 0
(5x-7)(2x+1) = 0
(5x-7) = 0 and (2x+1) = 0
x = 7/5, -1/2 Ans.
Que-16: 15x²-28 = x
Solution- 15x²-28 = x
15x²-x-28 = 0
15x²+20x-21x-28 = 0
5x(3x+4)-7(3x+4) = 0
(5x-7)(3x+4) = 0
(5x-7) = 0 and (3x+4) = 0
x = 7/5, -4/3 Ans.
Que-17: 8x²+15 = 26x
Solution- 8x²+15 = 26x
8x²-26x+15 = 0
8x²-6x-20x+15 = 0
2x(4x-3)-5(4x-3) = 0
(2x-5)(4x-3) = 0
(2x-5) = 0 and (4x-3) = 0
x = 5/2, 3/4 Ans.
Que-18: 3x²+8 = 10x
Solution- 3x²+8 = 10x
3x²-10x+8 = 0
3x²-6x-4x+8 = 0
3x(x-2)-4(x-2) = 0
(3x-4)(x-2) = 0
(3x-4) = 0 and (x-2) = 0
x = 4/3, 2 Ans.
Que-19: x(6x-11) = 35
Solution- x(6x-11) = 35
6x²-11x-35 = 0
6x²+10x-21x-35 = 0
2x(3x+5)-7(3x+5) = 0
(2x-7)(3x+5) = 0
(2x-7) = 0 and (3x+5) = 0
x = 7/2, -5/3 Ans.
Que-20: 6x(3x-7) = 7(7-3x)
Solution- 6x(3x-7) = 7(7-3x)
18x²-42x = 49-21x
18x²-42x+21x-49 = 0
6x(3x-7)+7(3x-7) = 0
(6x+7)(3x-7) = 0
(6x+7) = 0 and (3x-7) = 0
x = -7/6, 7/3 Ans.
Que-21: 2x²-9x+10 = 0, when (i) x ∈ N (ii) x ∈ Q
Solution- 2x²-9x+10 = 0
2x²-4x-5x+10 = 0
2x(x-2)-5(x-2) = 0
(2x-5)(x-2) = 0
(2x-5) = 0 and (x-2) =
x = 2, 5/2
(i) x ∈ N
x = 2 Ans.
(ii) x ∈ Q
x = 2, 5/2 Ans.
Que-22: 4x²-9x-100 = 0, when x ∈ Q
Solution- 4x²-9x-100 = 0
4x²+16x-25x-100 = 0
4x(x+4)-25(x+4) = 0
(4x-25)(x+4) = 0
(4x-25) = 0 and (x+4) = 0
x = 25/4, -4
when x ∈ Q
x = 25/4, -4 Ans.
Que-23: 3x²+11x+10 = 0, when x ∈ I
Solution- 3x²+11x+10 = 0
3x²+6x+5x+10 = 0
3x(x+2)+5(x+2) = 0
(3x+5)(x+2) = 0
(3x+5) = 0 and (x+2) = 0
x = -5/3, -2
when x ∈ I
x = -2 Ans.
Que-24: x+(1/x) = 3*(1/3), x ≠ 0
Solution- x+(1/x) = 3*(1/3)
(x²+1)/x = 10/3
3x²+3 = 10x
3x²-10x+3 = 0
3x²-9x-x+3 = 0
3x(x-3)-1(x-3) = 0
(3x-1)(x-3) = 0
(3x-1) = 0 and (x-3) = 0
x = 1/3, 3 Ans
Que-25: 5x-(35/x) = 18
Solution- 5x-(35/x) = 18
(5x²-35)/x = 18
5x²-35 = 18x
5x²-18x-35 = 0
5x²+7x-25x-35 = 0
x(5x+7)-5(5x+7) = 0
(x-5)(5x+7) = 0
(x-5) = 0 and (5x+7) = 0
x = 5, -7/5 Ans.
Que-26: 10x-(1/x) = 3
Solution- 10x-(1/x) = 3
(10x²-1)/x = 3
10x²-1 = 3x
10x²-3x-1 = 0
10x²+2x-5x-1 = 0
2x(5x+1)-1(5x+1) = 0
(2x-1)(5x+1) = 0
(2x-1) = 0 and (5x+1) = 0
x = 1/2, -1/5 Ans.
Que-27: 3a²x²+8abx+4b² = 0, a ≠ 0
Solution-In 3a²x²+8abx+4b² = 0
Roots are [−b±√(b²−4ac)]/2a
Roots are = [−8ab±√[{(8ab)²−4×3a²×4b²}]/2×3a²
Therefore, [−8ab±√64a²b² − 48a²b²]/6a²
⇒ [−8ab±4ab]/6a²
If [−8ab − 4ab]/6a² = [−12ab]/6a² = −2b/a
Or [−8ab + 4ab]/6a² = [−4ab]/6a² = −2b/3a
Then roots are −2b/a and −2b/3a Ans.
Que-28: 4x²-4ax+(a²-b²) = 0, where a,b ∈ R
Solution- 4x2 – 4ax + (a2 – b2) = 0 where a , b ∈ R
⇒ 4x2 – {2(a + b)x + 2(a – b)x} + a2 – b2 = 0
⇒ {4x2 – 2(a + b)x} – {2(a – b)x – (a2 – b2)} = 0
⇒ 2x{2x – (a + b)} – (a – b) {2x – (a + b)} = 0
⇒ {2x – (a + b)} {2x – (a – b)} = 0
⇒ 2x – (a + b) = 0
or
2x – (a – b) = 0
x = (a+b)/2 or x = (a-b)/2 Ans.
Que-29: 5x²-12x-9 = 0, when (i) x ∈ I (ii) x ∈ Q
Solution- 5x²-12x-9 = 0
5x²+3x-15x-9 = 0
x(5x+3)-3(5x+3) = 0
(x-3)(5x+3) = 0
(x-3) = 0 and (5x+3) = 0
x = 3, -3/5
(i) when x ∈ I
x = 3 Ans.
(ii) when x ∈ Q
x = 3, -3/5 Ans.
Que-30: 2x²-11x+15 = 0, when (i) x ∈ N (ii) x ∈ I
Solution- 2x²-11x+15 = 0
2x²-6x-5x+15 = 0
2x(x-3)-5(x-3) = 0
(2x-5)(x-3) = 0
(2x-5) = 0 and (x-3) = 0
x = 5/2, 3
(i) x ∈ N
x = 3 Ans.
(ii) x ∈ I
x = 3 Ans.
Que-31: √3x²+11x+6√3 = 0
Solution- √3x² + 11x + 6√3 = 0
⇒ √3x² + 9x + 2x + 6√3 = 0
⇒ √3x(x + 3√3) + 2(x + 3√3) = 0
⇒ (x + 3√3) (√3x + 2) = 0
⇒ (√3x + 2) = 0, (x + 3√3) = 0
⇒ x = -2/√3, x = -3√3 Ans.
Que-32: 2√5x²-3x-√5 = 0
Solution-2√5x²−3x−√5 = 0
⇒ 2√5x²+2x−5x-√5 = 0
⇒ 2x(√5x+1)−√5(√5x+1)
⇒(√5x+1)(2x−√5) = 0
⇒√5x+1 = 0 or 2x−√5 = 10
⇒x = −1√5 or x = √5/2 Ans.
Que-33: x²-(1+√2)x+√2 = 0
Solution- x²-(1+√2)x+√2 = 0
x²-x-√2x+√2 = 0
x(x-1)√2(x-1) = 0
(x-1)(x-√2) = 0
(x-1) = 0 or (x-√2) = 0
x = 1, √2 Ans.
Que-34: (x+1)/(x-1) = (3x-7)/(2x-5)
Solution- (x+1)/(x-1) = (3x-7)/(2x-5)
After cross multiplying , we get
(x+1) (2x-5) = (3x-7) (x-1)
2x2 -3x -5 = 3x2 – 10x + 7
x²- 7x + 12 = 0
(x-4) ( x-3 ) = 0
x = 4 or x = 3 Ans.
Que-35: (3x+1)/(7x+1) = (5x+1)/(7x+5)
Solution- (3x+1)/(7x+1) = (5x+1)/(7x+5)
(3x+1)(7x+5) = (5x+1)(7x+1)
21x²+15x+7x+5 = 35x²+5x+7x+1
35x²-21x²+12x-22x+1-5 = 0
14x²-10x-4 = 0
7x²-5x-2 = 0
7x²+2x-7x-2 = 0
x(7x+2)-1(7x+2) = 0
(7x+2)(x-1) = 0
(7x+2) = 0 or (x-1) = 0
x = -2/7, 1 Ans.
Que-36: 5/(2x+1) + 6/(x+1) = 3
Solution- 5/(2x+1) + 6/(x+1) = 3
[5x+5+12x+6]/[(2x+1)(x+1)] = 3
17x+11 = (3x+3)(2x+1)
17x+11 = 6x²+3x+6x+3
17x+11 = 6x²+9x+3
6x²−8x−8 = 0
6x²−12x+4x−8 = 0
6x(x−2)+4(x−2) = 0
(x−2)(6x+4) = 0
x = 2,−4/6
x = 2, -2/3 Ans.
Que-37: 2x/(x-4) + (2x-5)/(x-3) = 25/3
Solution- 2x/(x-4) + (2x-5)/(x-3) = 25/3
6x/(x-4) + (6x-15)/(x-3) = 25
[6x(x-3) + (6x-15)(x-4)]/[(x-3)(x-4)] = 25
6x²-18x+6x²-24x-15x+60 = 25(x-3)(x-4)
12x²-32x+60 = 25(x²-4x-3x+12)
12x²-32x+60 = 25x²-175x+300
25x²-12x²-175x+32x+300-60 = 0
13x²-118x+240 = 0
x²-118x/13+240/13 = 0
x²-6x-(40x/13)+(240/13) = 0
x(x-6)-(40/13)(x-6) = 0
(x-6)(x-40/13) = 0
x = 6, 40/13 Ans.
Que-38: (x+3)/(x-2) – (1-x)/x = 4*(1/4)
Solution-Consider the given equation.
(x+3)/(x−2) − (1−x)/x = 4*(1/4)
[x²+3x−x+x²+2−2x]/(x²−2x) = 17/4
4(2x²+2) = 17(x²−2x)
8x²+8 = 17x²−34x
9x²−34x−8 = 0
9x²+2x-36x-8 = 0
x(9x+2)-4(9x+2) = 0
(x-4)(9x+2) = 0
x = 4, −2/9 Ans.
Que-39: 1/(x-2) + 2/(x-1) = 6/x
Solution- 1/(x−2)+2/(x−1) = 6/x
= [x−1+2x−4]/[(x−2)(x−1)] = 6/x
= x(3x−5) = 6(x−2)(x−1)
= 3x²−5x = 6(x²−x−2x+2)
= 3x²−5x = 6x²−18x+12
= 13x−3x²−12 = 0
= 3x²−13x+12 = 0
= 3x²−4x−9x+12 = 0
= x(3x−4)−3(3x−4) = 0
(3x−4)(x−3)
so, x = 3 or x = 4/3 Ans.
Que-40: 2[x/(x+1)]² – 5[x/(x+1)] + 2 = 0, x ≠ -1
Solution- 2[x/(x+1)]² – 5[x/(x+1)] + 2 = 0
Let (x/(x+1)) = y
2y²-5y+2 = 0
2y²-4y-y+2 = 0
2y(y-2)-1(y-2) = 0
(2y-1)(y-2) = 0
(2y-1) = 0 or (y-2) = 0
y = 1/2 or y = 2
Put the value of y from the above equation
x/(x+1) = 1/2 or x/(x+1) = 2
2x = x+1 or x = 2x+2
2x-x = 1 or 2x-x = -2
x = 1 or x = -2 Ans.
Que-41: 5(3x+1)² + 6(3x+1) – 8 = 0
Solution- 5(9x²+1+6x) + 18x+6 – 8 = 0
45x²+5+30x+18x-2 = 0
45x²+48x+3 = 0
45x²+45x+3x+3 = 0
45x(x+1)+3(x+1) = 0
(45x+3)(x+1) = 0
x = -1, -3/45
x = -1, -1/15 Ans.
Que-42: √(x+15) = (x+3)
Solution- Squaring on both sides
x + 15 = (x + 3)2
⇒ x2 + 6x + 9 – x – 15 = 0
⇒ x2 + 5x – 6 = 0
⇒ x2 + 6x – x – 6 = 0
⇒ x(x + 6) –1(x + 6) = 0
⇒ (x + 6)(x – 1) = 0
Either x + 6 = 0,
then x = -6
or
x – 1 = 0,
then x = 1
∴ x = –6, 1
x = -6 is not a root
So, x = 1 Ans.
Que-43: √(2x+9) = (13-x)
Solution-√(2x+9) = 13−x
Squaring both sides
2x+9 = (13−x)²
2x+9 = 169−26x+x²
x²−28x−160 = 0
x²−20x−8x−160 = 0
x(x-20) -8(x-20)
(x-8)(x-20)
So, x = 8 and 20
x = 20 is not a root
So, x = 8 Ans.
Que-44: √(3x²-2) = (2x-1)
Solution- On squaring both sides, we get
3x2 – 2 = 4x2 + 1 – 4x
⇒ -x2 + 4x – 3 = 0
⇒ x2 – 4x + 3 = 0
⇒ x2 – 3x – x + 3 = 0
⇒ x(x – 3) -1(x – 3) = 0
⇒ (x – 3) (x – 1) = 0
⇒ x = 3 or x = 1
Hence, the solutions are [3, 1] Ans.
Que-45: √(3x²+x+5) = (x-3)
Solution-√(3x²+x+5) = (x-3)
Given equation:
3x²+x+5 = (x−3)²
⇒3x²+x+5 = x²−6x+9
⇒2x²+7x−4 = 0
⇒(x+4)(2x−1) = 0
⇒x = −4, 1/2 Ans.
Que-46: Find the quadratic equation whose solution set is :
(i) {2.-3} (ii) {-3,2/5} (iii) {2/5,-1/2}
Solution- (i) Since solution set is {2,-3)
⇒ x = 2 or x = -3
⇒ x – 2 = 0 or x + 3 = 0
⇒(x – 2)(x + 3) = 0
⇒ x2 + 3x – 2x – 6 = 0
⇒ x2 + x – 6 = 0 which is the required equation.
(ii) Since solution set is {-3,2/5}
x = -3 or x = 2/5
x + 3 = 0 Or 5x – 2 = 0
⟹ (x + 3) (5x – 2) = 0
5x²-2x+15x-6 = 0
5x²+13x-6 = 0 is the required solution.
(iii) Since solution set is {2/5,-1/2}
x = 2/5 or x = -1/2
5x-2 = 0 or 2x+1 = 0
(5x-2)(2x+1) = 0
10x²+5x-4x-2 = 0
10x² +x-2 = 0 is the required solution.
Que-47: Find the value of k for which x = 3 is a solution of the quadratic equation (k+2)x²-kx+6 = 0. Thus, find the other root of the equation.
Solution- Given : (k+2)x²−kx+6 = 0
Putting x = 3
(k+2)×9−k×3+6 = 0
9k+18−3k+6 = 0
6k = −24
k = −4 Ans.
Putting k = −4 in given equation
−2x²+4x+6 = 0
x²−2x−3 = 0
x²−3x+x−3 = 0
x(x−3)+1(x−3) = 0
(x+1)(x−3) = 0
x+1 = 0 or x−3 = 0
x = −1 or x = 3
Other root of the equation, x = −1 Ans.
–: End of Quadratic Equations Class 10 Exe- 5A RS Aggarwal Goyal ICSE Solutions :–
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