Quadratic EquationHard
Question
The equation $(x - a)(x - a - b) = 1$, where $a,b$ are positive constants has
Options
A.One root less than a and the other is greater than a.
B.One root less than $a + b$ and the other is greater than $a + b$.
C.One root less than a and the other is greater than $a + b$.
D.Roots lying between a and $a + b$.
Solution
$f(x) = (x - a)(x - a - b) - 1$
$$f(a) = f(a + b) = - 1 $$From graph it is clear that
root lie in internal $( - \infty,a) \cup (a + b,\infty)$
Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
If ratio of roots of the equations x2 + ax + b= 0 and x2 + px + q = 0 are equal, then -...Let $a,b \in C$. Let $\alpha,\beta$ be the roots of the equation $x^{2} + ax + b = 0$. If $\beta - \alpha = \sqrt{11}$ a...For the equation 3x2 + px + 3 = 0, p > 0, if one of the root is square ot the other, then p is equal to...If 1 lies between the roots of the quadratic equation 3x2 - 3sinθ.x + 2sin2θ = 2, then range of ′θ&...If α, β are roots of the equation px2 + qx − r = 0, then the value of is equal to -...