Quadratic EquationHard
Question
The sum of all the roots of the equation $(x - 1)^{2} - 5|x - 1| + 6 = 0$, is:
Options
A.4
B.3
C.1
D.5
Solution
Let $|x - 1| = t$
$${t^{2} - 5t + 6 = 0 }{t = 2\& t = 3 }{|x - 1| = 2\&|x - 1| = 3 }{x - 1 \pm 2\& x - 1 = \pm 3 }{x = 1 \pm 2\& x = 1 \pm 3 }$$∴ $root = 3, - 1,4, - 2$
∴ Sum of root $= 3 + ( - 1) + 4 + ( - 2) = 4$.
Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
Let ' p ' is a root of the equation $x^{2} - x - 3 = 0$. Then the value of $\frac{p^{3} + 1}{p^{5} - p^{4} - p^{3} + p^{...x2 + k ( 2x + 3) + 4 (x + 2) + 3k − 5 is a perfect square, if k equals -...If both roots of the quadratic equation $x^2 - 2kx + k^2 + k - 5 = 0$ are less than 5, then $k$ lies in the interval...If one of the factors of ax2 + bx + c and bx2 + cx + a is common, then -...For the roots of the equations 2x2 − 5x + 1 = 0 and x2 + 5x + 2 = 0 , which of the following statement is true -...