Determine the sum of imaginary roots of the equation $(2 x^{2} + x - 1) (4 x^{2} + 2 x - 3) = 6$

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Solution

Given $(2 x^{2} + x - 1) (4 x^{2} + 2 x - 3) = 6$
let $2 x^{2} + x = y$
$∴$ $(y - 1) (2 y - 3) = 6$
$\Rightarrow$ $2 y^{2} - 5 y - 3 = 0$
$\Rightarrow$ $y = 3, - \frac{1}{2}$
$∴$ $2 x^{2} + x - 3 = 0$
$2 x^{2} + x = - \frac{1}{2}$ $\Rightarrow$ $4 x^{2} + 2 x + 1 = 0$
Discriminants,
$D_{1} = 1 + 4 \times 3 = 13 > 0$ , real roots
$D_{2} = 4 - 16 = - 12 < 0$ , imaginary roots
$∴$ Sum of roots of $4 x^{2} + 2 x + 1 = 0$
$α + β = - \frac{2}{4} = - \frac{1}{2}$

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Which of the following equations has the sum of its roots as 3?
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Determine the sum of imaginary roots of the equation (2x2+x−1)(4x2+2x−3)=6

Given (2x2+x−1)(4x2+2x−3)=6let 2x2+x=y∴ (y−1)(2y−3)=6⇒ 2y2−5y−3=0⇒ y=3,−12∴ 2x2+x−3=02x2+x=−12 ⇒ 4x2+2x+1=0Discriminants,D1=1+4×3=13>0, real rootsD2=4−16=−12<0, imaginary roots∴ Sum of roots of 4x2+2x+1=0α+β=−24=−12

Determine the sum of imaginary roots of the equation $(2 x^{2} + x - 1) (4 x^{2} + 2 x - 3) = 6$