if the roots of the equation (b-c)x2+(c-a)x +a-b=0 are equal,then prove that 2b=a+c.help me !
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If the roots of quadratic equation are equal, discriminant is 0 .
b² - 4ac = 0
(c-a) ² - 4 ( b-c) (a - b ) = 0
c²+a²-2ac -4 { ab-b²-ac+bc} = 0
c²+a²-2ac-4ab+4b²+4ac-4bc= 0
a²+c²+(-2b)²+2(a)(c) +2(-2b)(a) +2(-2n)(c) = 0
( a + c -2b) ² = 0
a+ c -2b = 0
a+c=2b .
Hope helped!
b² - 4ac = 0
(c-a) ² - 4 ( b-c) (a - b ) = 0
c²+a²-2ac -4 { ab-b²-ac+bc} = 0
c²+a²-2ac-4ab+4b²+4ac-4bc= 0
a²+c²+(-2b)²+2(a)(c) +2(-2b)(a) +2(-2n)(c) = 0
( a + c -2b) ² = 0
a+ c -2b = 0
a+c=2b .
Hope helped!
Answered by
1
Heya User,
--> Considering that the roots are equal.. --> Δ [ Discriminant ] = 0
=> [ c - a ]² - 4 [ a - b ][ b - c ] = 0
=> [ c² - 2ac + a² ] - 4 [ ab - b² - ac + bc ] = 0
=> [ c² - 2ac + a² - 4ab + 4b² + 4ac - 4bc ] = 0
=> [ c² + a² - 4ab + 4b² + 2ac - 4bc ] = 0
=> [ a² + (-2b)² + c² + 2a(-b) + 2(-b)c + 2ac ] = 0
=> [ a - 2b + c ]² = 0
=> a - 2b + c = 0
=> a + c = 2b ----> ^_^ And we're done...
--> Considering that the roots are equal.. --> Δ [ Discriminant ] = 0
=> [ c - a ]² - 4 [ a - b ][ b - c ] = 0
=> [ c² - 2ac + a² ] - 4 [ ab - b² - ac + bc ] = 0
=> [ c² - 2ac + a² - 4ab + 4b² + 4ac - 4bc ] = 0
=> [ c² + a² - 4ab + 4b² + 2ac - 4bc ] = 0
=> [ a² + (-2b)² + c² + 2a(-b) + 2(-b)c + 2ac ] = 0
=> [ a - 2b + c ]² = 0
=> a - 2b + c = 0
=> a + c = 2b ----> ^_^ And we're done...
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