If α and β are the roots of the equation, 7 x2-3 x-2=0, then the value of (α/1-α2)+(β/1-β2) is equal to :

Question

If α and β are the roots of the equation, 7 x2-3 x-2=0, then the value of (α/1-α2)+(β/1-β2) is equal to : (A) (27/16) (B) (1/24) (C) (27/32) (D) (3/8)

Tardigrade Admin · Accepted Answer

7 x2-3 x-2=0 α+β=(3/7) α β=(-2/7) =(α/1-α2)+(β/1-β2)=(α+β-α β(α+β)/1-α2-β2+α2 β2) =((3/7)+(2/7)((3/7))/1-(α+β)2+2 α β+α2 β2) =(27/16)

Q. If $α$ and $β$ are the roots of the equation, $7 x^{2} - 3 x - 2 = 0,$ then the value of $\frac{α}{1 - α ^{2}} + \frac{β}{1 - β ^{2}}$ is equal to :

$\frac{27}{16}$

$\frac{1}{24}$

$\frac{27}{32}$

$\frac{3}{8}$

Q. If α and β are the roots of the equation, 7x2−3x−2=0, then the value of 1−α2α​+1−β2β​ is equal to :

1627​

241​

3227​

83​

7x2−3x−2=0 α+β=73​ αβ=7−2​ =1−α2α​+1−β2β​=1−α2−β2+α2β2α+β−αβ(α+β)​ =1−(α+β)2+2αβ+α2β273​+72​(73​)​ =1627​

Q. If $α$ and $β$ are the roots of the equation, $7 x^{2} - 3 x - 2 = 0,$ then the value of $\frac{α}{1 - α ^{2}} + \frac{β}{1 - β ^{2}}$ is equal to :

$\frac{27}{16}$

$\frac{1}{24}$

$\frac{27}{32}$

$\frac{3}{8}$