find all the zeroes of 2x4- 3x3+ 6x - 2 if it is given that two of its zeroes are - Maths - Real Numbers

Question

find all the zeroes of 2x4- 3x3+ 6x - 2 if
it is given that two of its zeroes are minus root 2 and root 2

Ajanta Trivedi · Accepted Answer

let the given polynomial be $f (x) = 2 x^{4} - 3 x^{3} + 6 x - 2$

if two roots are $- \sqrt{2} a n d \sqrt{2}$ .

let the other two roots be a and b.

sum of the roots:

$a + b - \sqrt{2} + \sqrt{2} = - \frac{- 3}{2} a + b = \frac{3}{2} ......... (1)$

product of the roots:

$a * b * (- \sqrt{2}) * (\sqrt{2}) = - \frac{- 2}{2} a b * (- 2) = 1 a b = - \frac{1}{2} ............ (2)$

now applying the identity:

$(a - b)^{2} = (a + b)^{2} - 4 a b (a - b)^{2} = (\frac{3}{2})^{2} - 4 * (- \frac{1}{2}) (a - b)^{2} = \frac{9}{4} + 2 = \frac{17}{4} a - b = \frac{\sqrt{17}}{2} ............ (3) [f o r a > b]$

adding eq(1) and eq(3):

$2 a = \frac{3}{2} + \frac{\sqrt{17}}{2} \Rightarrow a = \frac{3}{4} + \frac{\sqrt{17}}{4}$

from eq(1):

$b = \frac{3}{2} - \frac{3}{4} - \frac{\sqrt{17}}{4} b = \frac{3}{4} - \frac{\sqrt{17}}{4}$

thus the other zeroes of the polynomial is $\frac{3}{4} + \frac{\sqrt{17}}{4} a n d \frac{3}{4} - \frac{\sqrt{17}}{4}$

hope this helps you.

cheers!!

find all the zeroes of 2x4- 3x3+ 6x - 2 if it is given that two of its zeroes are minus root 2 and root 2.

find all the zeroes of 2x⁴- 3x³+ 6x - 2 if it is given that two of its zeroes are minus root 2 and root 2.