Step-by-step solution for: (√2+√10)^2

1 answer:

azamat3 years ago

5 0

Note: √a * √a = a
   √a * √b = √ab

(√2 + √10)² = (√2 + √10)(√2 + √10)
= √2(√2 + √10) + √10(√2 + √10)
= √2*√2 + √2*√10 + √10*√2 + √10*√10
=    2 + √20 + √20 + 10
      = (2 + 10) + (√20 + √20)
      = 12 + 2√20

√20 = √(4 *5) = √4 * √5 = 2√5

   = 12 + 2√20 = 12 + 2(2√5)
   = 12 + 4√5

You might be interested in

Help me pls! I have to get his done.<br>Factorise this!<br>2x^2+5x+3

MArishka [77]

Hello there!☺

$Answer:\boxed{(2x+3)(x+1)}$

$Explanation:$

$2x^2+5x+3$ Factorized

To factor $2x^2+5x+3$ , we will have to do it by grouping.

$2x^2+5x+3\\(2x+3)(x+1)$

$(2x+3)(x+1)$ is your answer.

Hope this helps!☺

8 0

3 years ago

27.36 divided by 3.14 use compatible numbers to estimate the quotient

stiks02 [169]

Answer:

Step-by-step explanation:

you can round the numbers

27/3= 9

27.36 rounds down to 27 and 3.14 rounds to 3

6 0

3 years ago

Read 2 more answers

What are the zeros of the polynomial equation f(x)=x^2-12x+20?

bagirrra123 [75]

$f(x)=x^2-12x+20\\\\\text{The zeros:}\\\\f(x)=0\to x^2-12x+20=0\\\\x^2-10x-2x+20=0\\\\x(x-10)-2(x-10)=0\\\\(x-10)(x-2)=0\iff x-10=0\ \vee\ x-2=0\\\\\boxed{x=10\ \vee\ x=2}$