If one root of bi-quadratic equation x4-2 x3-16 x2-22 x-7-0 is 2-3- Find the other three roots-A- -3-2-3 -3- 2-3-3 -3- -3-2-3 -2D- 2-3-3 -2

The correct option is D Since irrational roots occurs in pairs 2 √(3) must also be a root x = 2 + √(3) x 2 = √(3) x^2 4x + 4 = 3 x^2 4x + 1 = 0 Div ...

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Standard XI

Mathematics

Purely Real

If one root o...

Question

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.

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Solution

The correct option is C

Since irrational roots occurs in pairs

$2 - \sqrt{3}$ must also be a root

$x = 2 + \sqrt{3}$

$x - 2 = \sqrt{3}$

$x^{2} - 4 x + 4 = 3$

$x^{2} - 4 x + 1 = 0$

Divide $x^{4} + 2 x^{3} - 16 x^{2} - 22 x + 7 b y x^{2} - 4 x + 1$ we get the quotient is $x^{2} + 6 x + 7$
$∴ x^{2} + 6 x + 7 = 0 \Rightarrow x = - 3 \pm \sqrt{2}$

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If one root of bi-quadratic equation x4+2x3−16x2−22x+7=0is2+√3. Find the other three roots.

The correct option is C Since irrational roots occurs in pairs 2−√3 must also be a root x=2+√3 x−2=√3 x2−4x+4=3 x2−4x+1=0 Divide x4+2x3−16x2−22x+7 by x2−4x+1 we get the quotient is x2+6x+7 ∴x2+6x+7=0⇒x=−3±√2

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.