Question

find the product of 4 underoot 6 and 3 underoot 24. ​

Answer 1

Product of square roots

Given: two numbers 4√6 and 3√24

To find: the product of 4√6 and 3√24

Solution:

• Now, 4√6 = 4 × √6
• and 3√24 = 3 × √(2 × 2 × 2 × 3)
• = 3 × 2 × √(2 × 3)
• = 6 × √6

Before we find the required product, we must remember when we multiply two or more numbers under square roots can be multiplied under a single square root symbol.

• ∴ 4√6 × 3√24
• = 4 × √6 × 6 × √6
• = 4 × 6 × √6 × √6
• = 24 × √(6 × 6)
• = 24 × 6
• = 144

Answer: required product is 144.

Answer 2

QuestioN :

$\tt \bullet\int( 2x + 3 )^2 \,dx$

AnsWer :

4x²/3 + 9x + 6x² + c.

SolutioN :

Integration w.r.t.x

$\tt \mapsto I = \int( 2x + 3 )^2 \,dx$

$\tt \mapsto I = \int(4 {x}^{2} + 9 + 12x) \,dx$

★ By Sum rule of integration separate the terms.

$\tt \mapsto I = \int4 {x}^{2}\,dx + \int 9\,dx + \int 12x\,dx$

$\tt \mapsto I = 4\int {x}^{2}\,dx + 9\int \,dx + 12\int x\,dx$

$\tt \mapsto I = 4\frac{ {x}^{3}}{3} + 9x + \cancel{12}\frac{ {x}^{2} }{\cancel2} + c.$

$\tt \mapsto I = \frac{ {4x}^{3}}{3} + 9x + 6{x}^{2} + c.$

Therefore, the required value be 4x²/3 + 9x + 6x² + c.

MorE InformatioN.

$\tt \mapsto 1\int\,dx = x + c.$

$\tt \mapsto \int {x}^{n} \,dx = \frac{{x}^{n + 1}}{n + 1} + c.$

$\tt \mapsto \int\,Cosx\,dx = Sinx + c.$

$\tt \mapsto \int\,Sinx\,dx = -Cosx + c.$

$\tt \mapsto \int\,Sec^2x\,dx = tanx + c.$