Let a , b , c be in arithmetic progression. Let the centroid of the triangle with vertices ( a , c ) (2, b) and (a, b) be ((10/3), (7/3)). If α, β are the roots of the equation a x2+b x+1=0, then the value of α2+β2-α β is

Question

Let a , b , c be in arithmetic progression. Let the centroid of the triangle with vertices ( a , c ) (2, b) and (a, b) be ((10/3), (7/3)). If α, β are the roots of the equation a x2+b x+1=0, then the value of α2+β2-α β is (A) (71/256) (B) (69/256) (C) -(69/256) (D) -(71/256)

Tardigrade Admin · Accepted Answer

(a+2+a/3)=(10/3) a=4 and (c+b+b/3)=(7/3) c+2 b=7 also 2 b=a+c 2 b-a+2 b=7 b=(11/4) now 4 x2+(11/4) x+1=0 < αβ α2+β2-α β=(α+β)2-3 α β =((-11/16))2-3((1/4)) =(121/256)-(3/4)=(-71/256)

Q. Let $a, b, c$ be in arithmetic progression. Let the centroid of the triangle with vertices $(a, c)$ $(2, b)$ and $(a, b)$ be $(\frac{10}{3}, \frac{7}{3})$ . If $α, β$ are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the value of $α^{2} + β^{2} - α β$ is

$\frac{71}{256}$

$\frac{69}{256}$

$- \frac{69}{256}$

$- \frac{71}{256}$

Q. Let a,b,c be in arithmetic progression. Let the centroid of the triangle with vertices (a,c) (2,b) and (a,b) be (310​,37​). If α,β are the roots of the equation ax2+bx+1=0, then the value of α2+β2−αβ is

25671​

25669​

−25669​

−25671​

3a+2+a​=310​ a=4 and 3c+b+b​=37​ c+2b=7 also 2b=a+c 2b−a+2b=7 b=411​ now 4x2+411​x+1=0<βα​ α2+β2−αβ=(α+β)2−3αβ =(16−11​)2−3(41​) =256121​−43​=256−71​

Q. Let $a, b, c$ be in arithmetic progression. Let the centroid of the triangle with vertices $(a, c)$ $(2, b)$ and $(a, b)$ be $(\frac{10}{3}, \frac{7}{3})$ . If $α, β$ are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the value of $α^{2} + β^{2} - α β$ is

$\frac{71}{256}$

$\frac{69}{256}$

$- \frac{69}{256}$

$- \frac{71}{256}$