Home
Class 12
MATHS
If the quadratic equation a x^2+b x+6=0 ...

If the quadratic equation `a x^2+b x+6=0` does not have real roots and `b in R^+` , then prove that `a > m a x{(b^2)/(24),b-6}`

Text Solution

Verified by Experts

For the equation `ax^(2) + bx + 6 = 0` , roots are not real. Hence,
` D lt 0 rArr b^(2) - 24 a lt 0 or a gt (b^(2))/(24)`
Also, `f(0) = 6 gt 0 `. So ,
` f (-1) gt 0 rArr a + b + 6 gt 0`
`rArr a gt max {(b^(2))/(24), b -6}`.
Promotional Banner

Topper's Solved these Questions

  • THEORY OF EQUATIONS

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 2.13|9 Videos

  • THEORY OF EQUATIONS

    CENGAGE ENGLISH|Exercise Single Correct Answer Type : Exercise|89 Videos

  • THEORY OF EQUATIONS

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 2.11|8 Videos

  • STRAIGHT LINES

    CENGAGE ENGLISH|Exercise ARCHIVES (NUMERICAL VALUE TYPE)|1 Videos

  • THREE DIMENSIONAL GEOMETRY

    CENGAGE ENGLISH|Exercise All Questions|294 Videos

Similar Questions

Explore conceptually related problems

Root of the quadratic equation x^2+6x-2=0

If the equation x^2-b x+1=0 does not possess real roots, then range of b

The roots of the quadratic equation 2x^2 - x - 6 = 0 are

If a, b, c in R and the quadratic equation x^2 + (a + b) x + c = 0 has no real roots then

If c is positive and 2a x^2+3b x+5c=0 does not have any real roots, then prove that 2a-3b+5b>0.

If a x^2+b x+6=0 does not have distinct real roots, then find the least value of 3a+bdot

If a x^2+b x+6=0 does not have distinct real roots, then find the least value of 3a+bdot

If alpha is a real root of the quadratic equation a x^2+b x+c=0a n dbeta ils a real root of - a x^2+b x+c=0, then show that there is a root gamma of equation (a//2)x^2+b x+c=0 whilch lies between aa n dbetadot

If a, b in R and ax^2 + bx +6 = 0,a!= 0 does not have two distinct real roots, then :

If ax^(2)+bx+6=0 does not have two distinct real roots, a in R and b in R , then the least value of 3a+b is