IPMAT Indore 2024 · Quantitative Ability · Question 25

Step 1 — centre the substitution. The two brackets, {x}^{2}-5 and {x}^{2}-7, are symmetric about {x}^{2}-6. Put

u={x}^{2}-6 \Rightarrow{x}^{2}-5=u+1,{x}^{2}-7=u-1

Step 2 — expand using the symmetry. Odd powers cancel:

{\left( u+1 \right)}^{4}+{\left( u-1 \right)}^{4}=2\left({u}^{4}+6{u}^{2}+1 \right)

So the equation becomes

2\left({u}^{4}+6{u}^{2}+1 \right)=16 \Rightarrow{u}^{4}+6{u}^{2}-7=0

Step 3 — factorise as a quadratic in {u}^{2}.

\left({u}^{2}+7 \right)\left({u}^{2}-1 \right)=0

{u}^{2}=-7 is rejected (no real u), leaving {u}^{2}=1, i.e. u=\pm 1.

Step 4 — back-substitute.

u=1: {x}^{2}=7\Rightarrow x=\pm \sqrt{7}

u=-1: {x}^{2}=5\Rightarrow x=\pm \sqrt{5}

Step 5 — take the extremes. The real roots are -\sqrt{7},-\sqrt{5},\sqrt{5},\sqrt{7}.

\max -\min =\sqrt{7}-\left( -\sqrt{7}\right)=2\sqrt{7}

Answer: Option 2 — 2\sqrt{7}

Shortcut. Symmetry beats expansion. Whenever you see {\left( A \right)}^{n}+{\left( B \right)}^{n} with A and B equidistant from something, substitute the midpoint. Here u={x}^{2}-6 turns a quartic-in-disguise into {u}^{4}+6{u}^{2}-7=0, which factors on sight because -7=7\times \left( -1 \right).

Then the finish is pure structure: the root set is symmetric about 0, so the required difference is just twice the largest root, 2\sqrt{7}. Option (c) \sqrt{7} catches anyone who forgets the negative roots.