# Answering a question containing $\sqrt{1.5}$ without using calculator.

Of the following which is the best approximation of $\sqrt{1.5}(266)^{\frac{3}{2}}$?

(A)1,000

(B)2,700

(C)3,200

(D)4,100

(E)5,300

How can I answer this without using a calculator and in about 2.5 minutes?

$$\sqrt{\frac{3}{2} \times 266^3} = \sqrt{3 \times 133 \times 266^2} = 266 \times \sqrt{399} \simeq 266 \times 20 = 5320$$

The difference between $5320$ and the true answer is no more than $266$, since $\sqrt{399}$ is no more than $1$ away from $400$. (It's actually a much better approximation even than that.)

• +1 Your answer is better than mine as you can provably give an upper bound for the error. Apr 11, 2017 at 7:09

The given 4 alternatives are far from each other, and so we can calculate approximately choosing a convenient number close by.

(1) To calculate $266^{3/2}$, square root of 266 is needed. We approximate 266 to 256 as its square roots is nice $16$. Now to compensate for the reduction of this value we will increase $\sqrt{150/100}$ to $\sqrt{169/100}=1.3$.

So the final answer will be close to $1.3\times 16^3=1.3\times4096$ which should exceed 5000. So I will vote for the option E) $5300$

$\sqrt{1.5}(266)^{\frac{3}{2}} = 266 \times\sqrt{1.5 \times 266} = 266 \times\sqrt{266 + 133} = 266 \times\sqrt{399}$

Note that $\sqrt{399}$ is very close to $\sqrt{400} = 20$.

So $266 \times\sqrt{399} \approx 266 \times 20 = 5320.$

By the way, if you want a bit more accurracy, let $f(x) = \sqrt{x}$. Then $f'(x) = \dfrac{1}{2 \sqrt x}$

\begin{align} f(400+\delta) &\approx f(400) + \delta f'(400) \\ f(400-1) &\approx \sqrt{400} - \dfrac{1}{2 \sqrt{400}} \\ \sqrt{399} &\approx 20 - \dfrac{1}{40} \\ \hline 266 \times\sqrt{399} &\approx 266 \times \left(20-\dfrac{1}{40} \right) \\ &\approx 5320 - 6.65 \\ &\approx 5313.35 \end{align}

To ten digits, the correct answer is $5313.345839$.