Question

If a is equals to 14-6 root 5 then find the value of root a +1 by root a

Answer 1

Answer:

the answer you needed

is 4.06

Answer 2

$\sqrt{a} +\dfrac{1}{\sqrt{a}}$ = $\dfrac{15-3\sqrt{5}}{4}$

Step-by-step explanation:

We have,

a = 14 - 6$\sqrt{5}$

To find, $\sqrt{a} +\dfrac{1}{\sqrt{a}}$ = ?

∴ a = 14 - 6$\sqrt{5}$

⇒ a = 9 + 5 - 6$\sqrt{5}$

⇒ a = $3^2$ + $\sqrt{5} ^2$ - 2(3)($\sqrt{5}$)

Using the algebraic identity,

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

⇒ a = $(3-\sqrt{5})^2$

∴ $\sqrt{a} +\dfrac{1}{\sqrt{a}}$

= $\sqrt{(3-\sqrt{5})^2} +\dfrac{1}{\sqrt{(3-\sqrt{5})^2}}$

= $(3-\sqrt{5})+\dfrac{1}{(3-\sqrt{5})}$

= $(3-\sqrt{5})+\dfrac{1}{(3-\sqrt{5})}\times \dfrac{(3+\sqrt{5})}{(3+\sqrt{5})}$

= $(3-\sqrt{5})+\dfrac{(3+\sqrt{5})}{3^2-\sqrt{5}^2}$

Using the algebraic identity,

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

= $(3-\sqrt{5})+\dfrac{(3+\sqrt{5})}{9-5}$

= $(3-\sqrt{5})+\dfrac{(3+\sqrt{5})}{4}$

= $3+\dfrac{3}{4}+\dfrac{\sqrt{5} }{4} -\sqrt{5}$

= $\dfrac{15}{4} -\dfrac{3\sqrt{5}}{4}$

= $\dfrac{15-3\sqrt{5}}{4}$

∴ $\sqrt{a} +\dfrac{1}{\sqrt{a}}$ = $\dfrac{15-3\sqrt{5}}{4}$