Home
Class 12
MATHS
If 2a+3b+6c = 0, then show that the equa...

If 2a+3b+6c = 0, then show that the equation `a x^2 + bx + c = 0` has atleast one real root between 0 to 1.

Text Solution

Verified by Experts

Given `2a+3b+6c=0`
`impliesa/3+b/2+c=0`…..i
Let `f'(x)=ax^(2)+bx+c`
Then `f(x)=(ax^(3))/3+(bx^(2))/2+cx+d`
Now `f(0)=d` and `f(1)=a/3+b/2+c+d`
`=0+d` [from Eq(i) ]
Since `f(x)` is a polynomial of three degree, then `f(x)` is continuous and differentiable everywhere and `f(0)=f(1)`, then by Rolle's theorem `f'(x)=0` i.e. `ax^(2)+bx+c=0` has atleast one real root between 0 and 1.
Promotional Banner

Similar Questions

Explore conceptually related problems

If a + b + c = 0 , then the equation equation : 3ax^(2) + 2bx + c = 0 has :

If 2 a+3 b+6 c=0, a, b, c in R then the equation a x^(2)+b x+c=0 has a root in

The equation 3x^(2)+4ax+b=0 has atleast one root in (0, 1) if :

If a+b+c=0 , then the equation 3ax^(2)+2bx+c=0 has :

Show that the equation 2(a^(2)+b^(2))x^(2)+2(a+b)x+1=0 has no real roots when a!=b .

If a gt 0, b gt 0 and c gt 0 , then both the roots of the equations ax^2 + bx + c =0

In the equations ax^(2) +bx +c=0 , if b=0 then the equations.

If 0 lt a lt b lt c and the roots alpha,beta of the equation ax^2 + bx + c = 0 are non-real complex numbers, then

In the equations ax^(2) + bx+ c =0 , if one roots is negative of the other then:

If the equation ax^(2) + 2bx - 3c = 0 has non-real roots and ((3c)/(4)) lt (a + b) , then c is always :