x²-(x+5)(x-5)/x-5=3/7
Answers
Answer:
=x (2)(x5)(x-5)÷x-5=3/7
=x (2) (5x)(x-5)=3/7÷x-5
7×x2×5x=3
35x (3)=3
x (3)=3/35
x=3÷3/35
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NCERT Solution for 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 – × + 1 = (x – 2)3
Sol. (i) (x + 1)2 = 2(x – 3)
We have:
(x + 1)2 = 2 (x – 3) x2 + 2x + 1 = 2x – 6
⇒ x2 + 2x + 1 – 2x + 6 = 0
⇒ x2 + 70
Since x2 + 7 is a quadratic polynomial
∴ (x + 1)2 = 2(x – 3) is a quadratic equation.
(ii) x2– 2x = (–2) (3 – x)
We have:
x2 – 2x = (– 2) (3 – x)
⇒ x2 – 2x = –6 + 2x
⇒ x2 – 2x – 2x + 6 = 0
⇒ x2 – 4x + 6 = 0
Since x2 – 4x + 6 is a quadratic polynomial
∴ x2 – 2x = (–2) (3 – x) is a quadratic equation.
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
We have:
(x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x2 – x – 2 = x2 + 2x – 3
⇒ x2 – x – 2 – x2 – 2x + 3 = 0
⇒ –3x + 1 = 0
Since –3x + 1 is a linear polynomial
∴ (x – 2) (x + 1) = (x – 1) (x + 3) is not quadratic equation.
(iv) (x – 3) (2x + 1) = x(x + 5)
We have:
(x – 3) (2x + 1) = x(x + 5)
⇒ 2x2 + x – 6x – 3 = x2 + 5x
⇒ 2x2 – 5x – 3 – x2 – 5x – 0
⇒ x2 + 10x – 3 = 0
Since x2 + 10x – 3 is a quadratic polynomial
∴ (x – 3) (2x + 1) = x(x + 5) is a quadratic equation.
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
We have:
(2x – 1) (x – 3) = (x + 5) (x – 1)
⇒ 2x2 – 6x – x + 3 = x2 – x + 5x – 5
⇒ 2x2 – x2 – 6x – x + x – 5x + 3 + 5 = 0
⇒ x2 – 11x + 8 = 0
Since x2 – 11x + 8 is a quadratic polynomial
∴ (2x – 1) (x – 3) = (x + 5) (x – 1) is a quadratic equation.
(vi) x2 + 3x + 1 = (x – 2)2
We have:
x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = x2 – 4x + 4
⇒ x2 + 3x + 1 – x2 + 4x – 4 =0
⇒ 7x – 3 = 0
Since 7x – 3 is a linear polynomial.
∴ x2 + 3x + 1 = (x – 2)2 is not a quadratic equation.
(vii) (x + 2)3 = 2x(x2 – 1)
We have:
(x + 2)3 = 2x(x2 – 1)
x3 + 3x2(2) + 3x(2)2 + (2)3 = 2x3 – 2x
⇒ x3 + 6x2 + 12x + 8 = 2x3 – 2x
⇒ x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0
⇒ –x3 + 6x2 + 14x + 8 = 0
Since –x3 + 6x2 + 14x + 8 is a polynomial of degree 3
∴ (x + 2)3 = 2x(x2 – 1) is not a quadratic equation.
(viii) x3 – 4x2 – x + 1 = (x – 2)3
We have:
x3 – 4x2 – x + 1 = (x – 2)3
⇒ x3 – 4x2 – x + 1 = x3 + 3x2(– 2) + 3x(– 2)2 + (– 2)3
⇒ x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8
⇒ x3 – 4x2 – x – 1 – x3 + 6x2 – 12x + 8 = 0
2x2 – 13x + 9 = 0
dth] = 80
2 [Length + x] 80
⇒ Length = (40 – x) metres
∴ Area of the rectangle = Length × breadth
= (40 – x) × x sq. m
= 40x – x2
Now, according to the given condition,
Area of the rectangle = 400 m2
∴ 40x – x2 = 400
⇒ – x2 + 40x – 400 = 0
⇒ x2 – 40x + 400 = 0
Comparing (1) with axe + bx + c = 0, we get
a = 1
b = –40
c = 400
∴ b2 – 4ac = ( – 40)2 – 4 (1) (400)
= 1600 – 1600 – 0
Thus, the equation (1) has two equal and real roots.
∴ Breadth, x = 20 m
∴ Length = (40 – x) = (40 – 20) m = 20 m.
Download Class 10 Math Arithmetic Progression NCERT Solutions as PDF
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Foundation Program For Class IX & X
STUDY MATERIAL FOR CBSE CLASS 10 MATH
Chapter 1 - Areas related to Cirles
Chapter 2 - Arithmetic Progression
Chapter 3 - Circles
Chapter 4 - Constructions
Chapter 5 - Coordinate Geometry
Chapter 6 - Introduction to Trigonometry
Chapter 7 - Linear Equations in two variables
Chapter 8 - Polynomials
Chapter 9 - Probability
Chapter 10 - Quadratic Equations
Chapter 11 - Real Numbers
Chapter 12 - Statistics
Chapter 13 - Surface Areas and Volumes
Chapter 14 - Triangles
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