Question

If a^2+9/a ^2=31 what is the value a-3/a

Answer 1

$a^2+ \frac{9}{a^2}=31\\ \\ \Rightarrow a^2+ \frac{9}{a^2}-6=31-6\\ \\ \Rightarrow (a)^2+ (\frac{3}{a})^2-2 \times a \times \frac{3}{a} =25\\ \\ \Rightarrow (a- \frac{3}{a})^2 =25\\ \\ \Rightarrow a- \frac{3}{a}= \pm \sqrt{25} = \pm 5$

Value of a-(3/a) can be 5 or -5.

Answer 2

$a-\dfrac{3}{a}=5$

Step-by-step explanation:

We have,

$a^2+\dfrac{9}{a^2} =31$

To find, $a-\dfrac{3}{a}=?$

Using identity,

$(a-b)^{2}=a^{2}+b^{2}-2ab$

∴ $(a-\dfrac{3}{a})^{2}=a^{2}+(\dfrac{3}{a})^{2}-2a(\dfrac{3}{a})$

⇒ $(a-\dfrac{3}{a})^{2}=a^{2}+\dfrac{9}{a^2}-2(3)$

⇒ $(a-\dfrac{3}{a})^{2}=a^{2}+\dfrac{9}{a^2}-6$     ....... (1)

Put $a^2+\dfrac{9}{a^2} =31$ in (1), we get

⇒ $(a-\dfrac{3}{a})^{2}=31-6$

⇒$(a-\dfrac{3}{a})^{2}=25$

⇒$(a-\dfrac{3}{a})^{2}=5^2$

⇒$a-\dfrac{3}{a}=5$

Hence, $a-\dfrac{3}{a}=5$