Question

Find the number of integral solutions of: \[|{x^2} - 1| + |{x^2} - 5x + 6| = |5x - 7|\].

Answer 1

Hint: Some basic formulae to solve such questions:
1.The property of modulus \[|a| + |b| \leqslant |a - b|\]
2.Use splitting the middle term method to solve quadratic equations.

Complete step-by-step answer:
We can clearly see that above equation is in the form of \[|a| + |b| \leqslant |a - b|\].
Where,
 \[|a| = |{x^2} - 1|\];
\[|b| = |{x^2} - 5x + 6|\];
\[|a - b| = |{x^2} - 1 - ({x^2} - 5x + 6)| \Rightarrow |a - b| = |5x - 7|\]
\[
   \Rightarrow ({x^2} - 1) + ({x^2} - 5x + 6) \leqslant (5x - 7) \\
   \Rightarrow 2{x^2} - 5x + 5 \leqslant 5x - 7 \\
   \Rightarrow 2{x^2} - 10x + 12 \leqslant 0 \\
   \Rightarrow {x^2} - 5x + 6 \leqslant 0 \\
   \Rightarrow {x^2} - 3x - 2x + 6 \leqslant 0 \\
   \Rightarrow x(x - 3) - 2(x - 3) \leqslant 0 \\
   \Rightarrow (x - 3)(x - 2) \leqslant 0 \\
   \Rightarrow 2 \leqslant x \leqslant 3 \\
\]
Required integrals solutions: \[2 \leqslant x \leqslant 3\].

Note: Two bracket linear equalities must be known to student’s to solve such kinds of questions.
A quadratic equation can at most give two solutions. The solutions can be real and distinct; real and equal or in complex form depending upon the value of D is greater than 0, equals 0 or less than 0.