Question

How do you evaluate $4\sqrt{112}+5\sqrt{56}-9\sqrt{126}$ ?

Answer 1

Hint: We know that the ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$ using this formula we can write $\sqrt{ab}$ is equal to $\sqrt{a}\sqrt{b}$
In the given formula we can see that in 112 , 56 , 126 we have a common factor that is 14
We can take the common factor from the equation and solve it.

Complete step-by-step answer:
We have to evaluate $4\sqrt{112}+5\sqrt{56}-9\sqrt{126}$
We know that $\sqrt{ab}$ is equal to $\sqrt{a}\sqrt{b}$ , we can apply this to $\sqrt{112}$
So we can write $\sqrt{112}$ as $\sqrt{14\times 8}$
So $\sqrt{112}=2\sqrt{2}\times \sqrt{14}$
Similarly we can say $\sqrt{56}=2\sqrt{14}$
And $\sqrt{126}=3\sqrt{14}$
Now putting the value of these term in the equation we get
$\Rightarrow 4\sqrt{112}+5\sqrt{56}-9\sqrt{126}=8\sqrt{2}\sqrt{14}+10\sqrt{14}-27\sqrt{14}$
If we take $\sqrt{14}$ common we will get
$\Rightarrow 4\sqrt{112}+5\sqrt{56}-9\sqrt{126}=\left( 8\sqrt{2}-17 \right)\sqrt{14}$
So the value of $4\sqrt{112}+5\sqrt{56}-9\sqrt{126}$ is equal to $\left( 8\sqrt{2}-17 \right)\sqrt{14}$

Note: The equation given in the question is negative we can see that the value of the equation is $\left( 8\sqrt{2}-17 \right)\sqrt{14}$ . if we compare $8\sqrt{2}$ and 17 we can say that 17 is greater than $8\sqrt{2}$. We can evaluate this by squaring both numbers. Square of $8\sqrt{2}$ is equal to 128 and square of 17 is 289. So 17 is greater than $8\sqrt{2}$. But in the case of a negative number the number if the square of one number is greater than the square of another, then the number is less than the other number. The square root of any number which is not a square number is an irrational number. So $\sqrt{14}$ is an irrational number and if we $\left( 8\sqrt{2}-17 \right)\sqrt{14}$ is also an irrational number.