- Rational Numbers
- Linear Equations in One Variable
- Understanding Quadrilaterals
- Practical Geometry
- Data Handling
- Squares and Square Roots
- Cubes and Cube Roots
- Comparing Quantities
- Algebraic Expressions and Identities
- Visualising Solid Shapes
- Mensuration
- Exponents and Powers
- Direct and Inverse Proportions
- Factorisation
- Introduction to Graphs
- Playing with Numbers
NCERT Solutions Class 8 Mathematics Solutions for Exercise 9.3 in Chapter 9 - Algebraic Expressions and Identities
Prev
Q4) (a) Simplify 3x\left(4x-5\right)+3 and find values for
(i) x = 3
(ii) x = \frac{1}{2}
(b) Simplify a\left(a^2+a+1\right)+5 and find its value for
(i) a = 0
(ii) a = 1
(iii) a = -1
Solution:
(a) 3x\left(4x-5\right)+3
= 3x\times4x-3x\times5+3
= 12x^2-15x+3
(i) For x = 3, 12x^2-15x+3
= 12\left(3\right)^2- 15 x 3 + 3
= 12 x 9 - 45 + 3
= 108 - 45 + 3
= 66
(ii) For x=\frac{1}{2},12x^2-15x+3
= 12\left(\frac{1}{2}\right)^2-15\times\frac{1}{2}+3
= 12\times\frac{1}{4}-\frac{15}{2}+3
= 6-\frac{15}{2}=\frac{12-15}{2}=\frac{-3}{2}
(b) a\left(a^2+a+1\right)+5
= a\times a^2+a\times a+a\times1+5
= a^3+a^2+a+5
(i) For a = 0, a^3+a^2+a+5
= \left(0\right)^3+\left(0\right)^2+\left(0\right)+5
= 0 + 0 + 0 + 5 = 5
(ii) For a = 1, a^3+a^2+a+5
= \left(1\right)^3+\left(1\right)^2+\left(1\right)+5
= 1 + 1 + 1 + 5 = 8
(iii) For a = -1, a^3+a^2+a+5
= \left(-1\right)^3+\left(-1\right)^2+\left(-1\right)+5
= -1+1-1+5
= -2+6
= 4
Facebook
Whatsapp
Copy Link
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
