1.

If `ax^(2) + bx + 10 = 0` does not have two distinct real roots, then the least value of 5a + b, isA. -3B. -2C. 3D. none of these

Answer» Correct Answer - B
It is given that `ax^(2) + bx + 10 = 0` does not have two distinct real roots.
`therefore" "b^(2) - 40a le 0 rArr a ge (b^(2))/(40)`
Let y = 5a + b. Then,
`y = 5 xx (b^(2))/(40)+b=(b^(2)+8b)/(8)=(1)/(8) (b^(2) + 8b)`
`rArr" "y=(1)/(8) (b+4)^(2) - 2 ge -2`
Hence, the least value of 5a + b is -2.


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