Question

if the roots of quadratic equation (b-c)x2+(c-a)x+(a-b)=0 has equal roots, then show that 2b=a+c​

Answer 1

$\begin{gathered}{\Huge{\textsf{\textbf{\underline{\underline{\purple{Answer:}}}}}}}\end{gathered}$

$\implies$following is the equation of 2b = a+c=

The given equation is (b-c) x² + (c-a)x+(a−b)=0

The roots of the given equation are equal,

D=b²-4ac0

(c-a)²-4(b-c)(a - b) = 0 (c² + a²-2ac)-4 (ab-b²-ca+bc) = 0

⇒c² + a² - 2ac-4ab+4b² + 4ac-4bc = 0 ⇒a² +46² +c²-4ab-4bc + 2ac = 0

⇒a²+(-28)² +c²+2xax(−2b)+2x(-26)xc+2xcxa=0

= (a +(-2b)+c)² = 0 ((x+y+z)² = x² + y² +z²+2xy +2yz+2zx)

⇒a-2b+c=0

⇒2b=a+c

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