Let a, b and c satisfy the system of equations a+2b+3c=6, 4a+5b+6c=12 and 6a+9b=4. If the roots of the equation a+b+cx^2-abcx+a^(-1)+b^(-1)+c^(-1)=0…

Question

Let a, b and c satisfy the system of equations a+2b+3c=6, 4a+5b+6c=12 and 6a+9b=4. If the roots of the equation a+b+cx^2-abcx+a^(-1)+b^(-1)+c^(-1)=0 are α and β, then 1/(α)+1/(β) is equal to

MARKS Content Team · Accepted Answer

Correct answer: (D) 100/243. Using cramer’s rule ∆= 1 2 3 4 5 6 6 9 0 =1-54-2-36+36 =-54+72+18=36 ∆_1= 6 2 3 12 5 6 4 9 0 =6-54-2-24+388 =-324+48+264=-12 ∆_2= 1 6 3 4 12 6 6 4 0 =1-24-6-36+3-56 =-24+216-168=24 ∆_3= 1 2 6 4 5 12 6 9 4 =1-88-2-56+66 =-88+112+36=60 ⇒a=(∆_1)/(∆)=-1/3,b=(∆_2)/(∆)=2/3,c=(∆_3)/(∆)=5/3 ⇒ equation is 2x^2+10/27x-9/10=0 ⇒540x^2+100x-243=0 α+β=-5/27,αβ=-9/20 1/(α)+1/(β)=(α+β)/(αβ)=(-5)/27-920=100/243

Let $a, b$ and $c$ satisfy the system of equations $a + 2 b + 3 c = 6, 4 a + 5 b + 6 c = 12$ and $6 a + 9 b = 4$ . If the roots of the equation $(a + b + c) x^{2} - a b c x + (a^{- 1} + b^{- 1} + c^{- 1}) = 0$ are $α$ and $β$ , then $\frac{1}{α} + \frac{1}{β}$ is equal to

Solution

More Determinants Questions

Let a, b and c satisfy the system of equations a+2b+3c=6, 4a+5b+6c=12 and 6a+9b=4. If the roots of the equation a+b+cx2-abcx+a-1+b-1+c-1=0 are α and β, then 1α+1β is equal to

Solution

More Determinants Questions

Let $a, b$ and $c$ satisfy the system of equations $a + 2 b + 3 c = 6, 4 a + 5 b + 6 c = 12$ and $6 a + 9 b = 4$ . If the roots of the equation $(a + b + c) x^{2} - a b c x + (a^{- 1} + b^{- 1} + c^{- 1}) = 0$ are $α$ and $β$ , then $\frac{1}{α} + \frac{1}{β}$ is equal to