Question

If \[|x - 6| > |{x^2} - 5x + 9|\] , then $1 < x < 3$ .

Answer 1

Hint: Use discriminants which is $\sqrt {{b^2} - 4ac} $ and try to visualise the graph.

Complete step-by-step answer:

First of all, We know that, a quadratic equation is given in standard form $a{x^2} + bx + c = 0$ then we can find its root using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , Where $\sqrt {{b^2} - 4ac} $ is called discriminant. Discriminant tells us about the roots of the equation. If it’s positive then roots are real and the curve will cut the x axis at two different points, if it’s negative then the roots are imaginary and the curve does not have the x axis at all. If it’s equal to zero then it has exactly one real root and curve cut x axis at exactly one point. Note that the discriminant of the parabola $P:{x^2} - 5x + 9 = 0$ is ${5^2} - 4 \times 9 = - 11 < 0$ . It means it doesn't have any real roots and the curve will be either subside or down side of the x axis. Since the coefficient of ${x^2}$ is 1. Which is positive then the curve will be above x axis. Now, we are interested in those x where, the graph of $|x - 6|$ lies above P. Now, observe that, the line $l:x - 6 = 0$ intersects P if and only if \[{x^2} - 5x + 9 = x - 6\] that is if ${x^2} - 6x + 15 = 0$ . Which again is a quadratic question so we can calculate its discriminant. ${6^2} - 4 \times 1 \times 15 = - 24 < 0$ . Since, the discriminant is negative. Hence l lies below $P{\text{ }}\forall {\text{ }}x \geqslant 6$.

On the other hand the line $r:6 - x = 0$ intersect P at two points because \[{x^2} - 5x + 9 = 6 - x \Rightarrow {x^2} - 4x + 3 = 0\] has two real and distinct roots. Which are

\[\

  {x^2} - 4x + 3 = 0 \\

   \Rightarrow {x^2} - (1 + 3)x + 3 = 0 \\

   \Rightarrow {x^2} - x - 3x + 3 = 0 \\

   \Rightarrow x(x - 1) - 3(x - 1) = 0 \\

   \Rightarrow (x - 3)(x - 1) = 0 \\

   \Rightarrow x = 1,3 \\

\]

Which means tha graph of $r:6 - x = 0$ lies above the graph of P precisely for $1 < x < 3$ .

Note: Here, we have used the fact that P will always lie above x axes. There is one more way to solve this problem by breaking down the whole problem into 4 cases, then solve it.