Question

⠀ [100points]★Maths challenge★

Solve this :-

★Find all the Solutions of the equation ?

${( {x}^{2} - 7x + 11)}^{ {x}^{2} - 13x + 42 } = 1$
Hint:- Total number of solutions/root of the equation are "6".

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Answer 1

(x² - 7x + 11)^(x² - 13x + 42) = 1

Since, anything raised to the power 0 = 1

Now since the bases are same, we can compare the powers.

=>

Now, factorise the LHS by splitting the middle term,

x² - 13x + 42 = 0

=> x² - 7x - 6x + 42 = 0

=> x(x - 7) - 6(x - 7) = 0

=> (x - 6)(x - 7) = 0

=> (x - 6) = 0 ÷ (x - 7)

=> (x - 6) = 0

=> x = 6

Similarly,

(x - 7)(x - 6) = 0

=> (x - 7) = 0 ÷ (x - 6)

=> (x - 7) = 0

=> x = 7

Hence, your answer is, x = 6 or x = 7

                                                   AND

Consider aⁿ,

If a≠±1 and if a≠0, then aⁿ is always 1 whenever n is 0.----(1)

if a = 1 then aⁿ is always 1....(2)

if a = -1 then n should be even for aⁿ to be 1----(3)

Consider (x²-7x+11)^(x²-13x+42) =1

Case 1

If x²-13x+42 = 0

⇒ x² - 6x -7x + 42 = 0

⇒ (x - 6)(x-7) = 0

⇒ x = 6 or x = 7

If x = 6, x² - 7x + 11 = 6² -7.6 + 11 ≠ 0

Hence 6 is one solution

If x = 7, x² - 7x + 11 = 7² -7.7 + 11 ≠ 0

Hence 7 is one solution

Case 2:

If x² - 7x + 11 = 1

⇒x² -7x + 10 = 0

⇒(x - 2)(x - 5) = 0

⇒ x = 2 or x = 5 are another 2 solutions

If  x² - 7x + 11 = -1

⇒x² -7x + 12 = 0

⇒ (x - 3)(x - 4) = 0

=> x = 3 or x = 4

If x = 3, x² - 13x + 42 = 3² - 13*3 + 42 = 12 which is even

Hence , x = 3 is solution as stated in (3).

If x = 4, x² - 13x + 42 = 4² - 13*4 + 42 = 6 which is even

Hence , x = 4 is solution as stated in (3).

Thus the solutions are 2,3,4,5,6 and 7.

Answer 2

As any number to power 0 is 1

and 1 to the power anything is 1

Take

x^2 - 13x + 42 = 0

x^2 - 6x -7x +42 = 0

x( x-6( -7( x-6)) = 0

( x-6) ( x-7) = 0

x= 6,7

So 2 solutions for this

Take

x^2 - 7x +11 = 1

x^2 - 7x +10= 0

x^2 - 2x -5x +10= 0

x( x-2) -5( x-2) = 0

x= 5,2

SO 2 solutions

Now

As x^2 - 7x +11 can be taken as -1 and x^2 -13x +42 as any even number

So

x^2 - 7x +11 = -1

x^2 - 7x +12 = 0

x^2 - 4x - 3x +12 = 0

x( x-4) - 3( x-4) = 0

x-3)( x-4) = 0

x= 3,4

Let's check whether x^2 -13x +42 is even for x= 3,4

put x= 3

(3)^2 -13(3) +42 = 9 -39 +42 = 12

So its even

now x= 4

(4)^2 -13(4) +42 = 16 -52 +42 = 16 -10 = 6

its also even


SO NO OF SOLUTIONS

2+2 +2 = 6 Solutions