Question

3√98 − 7/2 √ 1/2 pls tel ll l l l l l

Answer 1

To find

The value of $3\sqrt{98}-\dfrac{7}{2}\sqrt{\dfrac{1}{2}}$

Step-by-step explanation

Let us find the simplified form of $\sqrt{98}$ first.

$\therefore \sqrt{98}=\sqrt{2\times 7\times 7}=7\sqrt{2}$

• Here let us remember an important rule of indices: $\sqrt{2}\times\sqrt{2}=2$.

• However the above expression comes from: $a^{b}\times a^{c}=a^{b+c}$.

Now, $3\sqrt{98}-\dfrac{7}{2}\sqrt{\dfrac{1}{2}}$

$=3\times 7\sqrt{2}-\dfrac{7}{2\sqrt{2}}$

$=21\sqrt{2}-\dfrac{7}{2\sqrt{2}}$

$=\dfrac{21\sqrt{2}\times 2\sqrt{2}-7}{2\sqrt{2}}$

$=\dfrac{84-7}{2\sqrt{2}}$

$=\dfrac{77}{2\sqrt{2}}$

$=\dfrac{77\times\sqrt{2}}{2\sqrt{2}\times\sqrt{2}}$ [Rationalizing the denominator]

$=\dfrac{77\sqrt{2}}{4}$

$=\dfrac{77}{4}\sqrt{2}$

Answer:

$3\sqrt{98}-\dfrac{7}{2}\sqrt{\dfrac{1}{2}}=\dfrac{77}{4}\sqrt{2}$