Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

State true or false. 
$\sqrt { 3 } + \sqrt { 4 }$ is an rational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers.
$\Rightarrow$  All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction.

$\sqrt{3}=1.732$ is an irrational.
$\sqrt{4}=2$ is rational.
Now,
$\Rightarrow$  $ \sqrt{3}+\sqrt{4}=1.732+2$
                        $=3.732$
$3.732$ cannot be converted into fraction.
$\therefore$  $\sqrt{3}+\sqrt{4}$ is an irrational number.
$\therefore$  Given statement is false.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

$\sqrt{5}\left{(\sqrt{5}+1)^{50}-(\sqrt{5}-1)^{50}\right}$ is?

  1. An irrational number
  2. $0$

  3. A natural number

  4. A prime number
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This expression involves powers of (sqrt(5)+1) and (sqrt(5)-1). Using the binomial expansion, the irrational parts cancel out, leaving an irrational result due to the leading sqrt(5).

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Find x if $\dfrac{\sqrt{3x+1}+\sqrt{3x-6}}{\sqrt{3x+1}-\sqrt{3x-6}}=7$.

  1. $2$
  2. $5$
  3. $3$
  4. $7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac { \sqrt { 3x+1 } +\sqrt { 3x-6 }  }{ \sqrt { 3x+1 } -\sqrt { 3x-6 }  } =7$

Rotational give :-

$\dfrac { \left( 3x+1 \right) +\left( 3x-6 \right) +2\sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  }  }{ \left( 3x+1 \right) -\left( 3x-6 \right)  } =7$

$\Rightarrow 6x-5+2\sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  } =49$

$\Rightarrow 2\sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  } =-6x+54$

$\Rightarrow \sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  } =-3x+27$
which gives, $x=5$