if an equation ax^2 + Bx +6 = 0 does not have distinct real roots then ,the least value of 3a +b will be ?

Question

if an equation ax^2 + Bx +6 = 0 does not have distinct real roots then ,the least value of 3a +b will be ?

Answer 1

Akshay,

 

This one takes a little patience to reason through and requires you to know about the quadratic formula and the meaning of "distinct real roots."  Do you know much about those concepts already?

 

There's a part of the formula under the square root symbol, called the discriminant, that will tell you if any real roots exist for a quadratic equation.  That's what you'll need to focus on here: b² - 4ac, where a is the same a as in the equation you've written, b is the B you've written [go with little b instead of the capitalized b], and c is 6.  So we can rewrite it as: b² - 24a after substituting in for c.  

 

Imaginary roots will exist when this expression is negative.

Repeating roots will exist when this expression is exactly equal to zero.

Distinct real roots will exist when this expression is greater than zero.

 

Let me know whether you need more help than that to figure out the minimum values for b and a that you would then put into the new expression "3a+b," ok?  I'll be back on here later on this afternoon.  Cheers!

Answer 2

David is correct.  Following up, one realizes that b^2 = 24a at the solution.  Thus, a = b^2/24

 

Then, 3a + b = 3 * b^2/24 + b = b^2/8 + b

We minimize either by completing the square or taking the first derivative.

Completing the square, b^2/8 + b = 1/8(b^2 + 8b) = 1/8(b^2+8b+16) - 1/8(16) = 1/8(b+4)^2 - 2

This is minimized at b = -4.  a = b^2/24 = (-4)^2/24 = 2/3

 

As a check, note that 2/3 x^2 - 4 x + 6 = 2/3 (x^2 - 6x + 9) = 2/3 (x - 3)^2 has one solution, not two.