Solve the equation $x^{4} + 2 x^{3} - 16 x^{2} - 22 x + 7 = 0$ give that it has a root $2 + \sqrt{3}$ .

Open in App

Solution

As we know that irrational roots occur in pair
So, other root will be $2 - \sqrt{3}$

Now, Sum of roots $= \frac{c o f f i c i e n t o f x^{3}}{c o f f i c i e n t o f x^{4}} = \frac{- 2}{1}$

Product of Roots $= \frac{c o n s t a n t}{c o f f i c i e n t o f x^{4}} = \frac{7}{1}$

Now, $(2 + \sqrt{3}) (2 - \sqrt{3}) c d = 7 \Rightarrow c d = 7$ ............ $(1)$

$(2 + \sqrt{3}) + (2 - \sqrt{3}) + c + d = - 2 \Rightarrow c + d = - 6$ ........... $(2)$

Putting $d = \frac{7}{c}$ ........ from eq $(1)$
$c + \frac{7}{c} = - 6 \Rightarrow c^{2} + 6 c + 7 = 0 \Rightarrow c = - 3 + \sqrt{2}, d = - 3 - \sqrt{2}$

Therefore, solution of above bi quadratic equation are $- 3 + \sqrt{2}, - 3 - \sqrt{2}, 2 - \sqrt{3}, 2 + \sqrt{3}$

Suggest Corrections

Similar questions

If one root of bi-quadratic equation $x^{4} + 2 x^{3} - 16 x^{2} - 22 x + 7 = 0 i s 2 + \sqrt{3}$ . Find the other three roots.

x^{4} - 2 x^{3} - 21 x^{2} + 22 x + 40 = 0

x^{4} + x^{3} - 16 x^{2} - 4 x + 48 = 0

8

x^{4} - 4 x^{2} + 8 x + 35 = 0

2 + \sqrt{- 3} .

2 x^{3} - x^{2} - 22 x - 24 = 0

3 : 4

Join BYJU'S Learning Program