if the roots of the quadriatic equation are equal then show that
a=b=c.
3x^2-2(a+b+c)x+ab+bc+ac=0
Answers
Given equation says :
3 x² - 2 ( a + b + c ) x + ab + bc + ac = 0
This is another quadratic equation .
The standard model of quadratic equation is :
ax² + bx + c = 0
Comparing that equation with this one , we get :
a = 3
b = - 2 ( a + b + c )
c = ab + bc + ac
Whenever we are given that the roots of a particular equation are equal , then we must write b² = 4 ac
So :
( - 2 ( a + b + c ) )² = 4 ( 3 ( ab + bc + ca ) )
Use the expansion : ( a + b + c )² = a² + b² + c² + 2 ( ab + bc + ca )
⇒ 4 ( a² + b² + c² + 2 ab + 2 bc +2 ac ) = 12 ( ab + bc + ca )
⇒ 4 a² + 4 b² + 4 c² + 8 ab + 8 bc + 8 ac = 12 ab + 12 bc + 12 ac
⇒ 4 a² + 4 b² + 4 c² = 4 ab + 4 bc + 4 ac
⇒ a² + b² + c² = ab + bc + ac
⇒ a² + b² + c² - ab - bc - ac = 0
Multiply both sides by 2 :
⇒ 2 a² + 2 b² + 2 c² - 2 ab - 2 bc - 2 ac = 0
⇒ a² + b² - 2 ab + a² + c² - 2 ac + b² + c² - 2 bc = 0
⇒ ( a - b )² + ( a - c )² + ( b - c )² = 0
Since squares cannot be negative :
every term = 0 .
a - b = 0
⇒ a = b
b - c = 0
⇒ b = c
Hence a = b = c .
Proved !
Answer:
a = b = c
Step-by-step explanation:
Given Quadratic Equation is 3x² - 2(a + b + c)x + ab + bc + ac = 0.
On comparing with ax² + bx + c = 0, we get
a = 3, b = -2(a + b + c), c = ab + bc + ac.
Given that the equation has equal roots.
∴ b² - 4ac = 0
⇒ [-2(a + b + c)]² - 4(3)(ab + bc + ac) = 0
⇒ 4(a + b + c)² - 12(ab + bc + ac) = 0
⇒ 4a² + 4b² + 4c² + 8ab + 8bc + 8ac - 12ab - 12bc - 12ac = 0
⇒ 4a² + 4b² + 4c² - 4ab - 4bc - 4ac = 0
⇒ a² + b² + c² - ab - bc - ac = 0
⇒ a² + b² + c² = ab + bc + ac
Multiply both sides by '2', we get
⇒ 2(a² + b² + c²) = 2(ab + bc + ac)
⇒ 2a² + 2b² + 2c² = 2ab + 2bc + 2ac
⇒ a² + a² + b² + b² + c² + c² = 2ab + 2bc + 2ca
⇒ a² + b² - 2ab + b² + c² - 2bc + c² + a² - 2ca = 0
⇒ (a - b)² + (b - c)² + (c - a)² = 0
⇒ a - b = 0, b - c = 0, c - a = 0
⇒ a = b, b = c, c = a
⇒ a = b = c.
Hope it helps!