Introduction of Quadratic Equations Class 10 Concise Exe-5A ICSE Maths Selina Solutions

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Introduction of Quadratic Equations Class 10 Concise Exe-5A ICSE Maths Selina Solutions. In this article you would learn about basic of Quadratic equations with some example questions and their answer. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Introduction of Quadratic Equations Class 10 Concise Exe-5A ICSE Maths Selina Solutions

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Introduction of Quadratic Equations Class 10 Concise Exe-5A ICSE Maths Selina Solutions

Board ICSE
Publications Selina
Subject Maths
Class 10th
Chapter-5 Quadratic Equations
Writer R.K. Bansal
Exe-5A Introduction of Quadratic Equations.

Introduction of Quadratic Equations

Class 10 Concise Exe-5A ICSE Maths Selina Solutions Ch-5 Quadratic Equations
Que-1: Find which of the following equations are quadratic :
(i) (3x-1)² = 5(x+8)        (ii) 5x²-8x = -3(7-2x)
(iii) (x-4)(3x+1) = (3x-1)(x+2)       (iv) x²+5x-5 = (x-3)²
(v) 7x³-2x²+10 = (2x-5)²      (vi) (x-1)² + (x+2)² +3(x+1) = 0.

Sol: (i) (3x – 1)2 = 5(x + 8)
⇒ (9x2 – 6x + 1) = 5x + 40
⇒ 9x2 – 11x – 39 =0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.

(ii) 5x2 – 8x = -3(7 – 2x)
⇒ 5x2 – 8x = 6x – 21
⇒ 5x2 – 14x + 21 =0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.

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(iii) (x – 4)(3x + 1) = (3x – 1)(x +2)
⇒ 3x2 + x – 12x – 4 = 3x2 + 6x – x – 2
⇒ 16x + 2 =0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.

(iv) x2 + 5x – 5 = (x – 3)2
⇒ x2 + 5x – 5 = x2 – 6x + 9
⇒ 11x – 14 =0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.

(v) 7x3 – 2x2 + 10 = (2x – 5)2
⇒ 7x3 – 2x2 + 10 = 4x2 – 20x + 25
⇒ 7x3 – 6x2 + 20x – 15 = 0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.

(vi) (x – 1)2 + (x + 2)2 + 3(x +1) = 0
⇒ x2 – 2x + 1 + x2 + 4x + 4 + 3x + 3 = 0
⇒ 2x2 + 5x + 8 = 0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.

Que-2: (i) Is x = 5 a solution of the quadratic equation x2 – 2x – 15 = 0?

Sol: x2 – 2x – 15 = 0
For x = 5 to be solution of the given quadratic equation it should satisfy the equation.
So, substituting x = 5 in the given equation, we get
L.H.S = (5)2 – 2(5) – 15
= 25 – 10 – 15
= 0
= R.H.S
Hence, x = 5 is a solution of the quadratic equation x2 – 2x – 15 = 0.

(ii) Is x = -3 a solution of the quadratic equation 2x2 – 7x + 9 = 0?

Sol: 2x2 – 7x + 9 = 0
For x = -3 to be solution of the given quadratic equation it should satisfy the equation
So, substituting x = 5 in the given equation, we get
L.H.S =2(-3)2 – 7(-3) + 9
= 18 + 21 + 9
= 48
≠ R.H.S
Hence, x = -3 is not a solution of the quadratic equation 2x2 – 7x + 9 = 0.

Que-3: If √(2/3) is a solution of equation 3x2 + mx + 2 = 0, find the value of m.

Sol: For x = √(2/3) to be solution of the given quadratic equation it should satisfy the equation
So, substituting x = √(2/3) in the given equation, we get
3{√(2/3)}² + m{√(2/3)} + 2 = 0
3(2/3) + m{√(2/3)} + 2 = 0
m = -4 × √(3/2) = -2√6
m = -2√6

Que-4: 2/3 and 1 are the solutions of equation mx2 + nx + 6 = 0. Find the values of m and n.

Sol: For x = 2/3 and x = 1 to be solutions of the given quadratic equation it should satisfy the equation
So, substituting x = 2/3 and x = 1 in the given equation, we get
m(2/3)² + n(2/3) + 6 = 0    and    m(1)² + n(1) + 6 = 0
m(4/9) + n(2/3) + 6 = 0   and   m + n + 6 = 0
4m + 6n + 54 = 0… (1)   and   m + n + 6 = 0 ……….. (2)
Solving equations (1) and (2) simultaneously,
4m  + 6n + 54 = 0 …..(1)
m + n  + 6 = 0 ….(2)
(1) – (2) × 6
⇒ -2m + 18 = 0
⇒ m = 9
Substitute in (2)
⇒ n = -15

Que-5: If 3 and -3 are the solutions of equation ax2 + bx – 9 = 0. Find the values of a and b.

Sol: For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the equation
So, substituting x = 3 and x = -3 in the given equation, we get
a(3)² + b(3) – 9 = 0   and  a(-3)² + b(-3) – 9 = 0
a(9) + 3b – 9 = 0   and   a(9) – 3b – 9 = 0
9a + 3b – 9 = 0 ……. (1)   and    9a – 3b – 9 = 0 ……… (2)
Solving equations (1) and (2) simultaneously,
9a + 3b – 9 = 0 …(1)
9a – 3b – 9 = 0 …(2)
(1) + (2)
⇒ 18a – 18 = 0
⇒ a = 1
Substitute in (2)
⇒ b = 0

— : End of Introduction of Quadratic Equations Class 10 Concise Exe-5A ICSE Maths Selina Solutions. :–

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