|-x²+3x-5| + x²-3x+5
Topic - Modulus Equations
Answer - ϕ
Juz' explain me how can it be ϕ
shouldn't it be like x∈R !?
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Answers
Given :-
| -x² + 3x - 5 | + x² - 3x + 5
To Prove :-
x ∈ ϕ
Used Concepts :-
- | -a | = a
- Value of iota ( i ) i.e √-1.
- For solutions of a quadratic polynomial ax² + bx + c it must be equal to 0 .
Solution :-
| -x² + 3x - 5 | + x² - 3x + 5
Let us assume that x² - 3x + 5 = y .
=> | - ( x² + 3x + 5 ) | + y
=> | -y | + y
=> y + y [ ∵ | -a | = a ]
=> 2y
But ,y = x² - 3x + 5
∴ 2y = 2 × ( x² - 3x + 5 )
=> 2x² - 6x + 10
Let , us assume that p ( x ) = 2x² - 6x + 10 .
So , For the values of " x " p ( x ) = 0
∴ 2x² - 6x + 10 = 0
=> Here it is a quadratic equation .
Where , a = 2 , b = -6 and c = 10
∴ D = b² - 4ac => D = ( -6 )² - 4 × 2 × 10
=> D = 36 - 80 = -44 => √D = √-44 = 2√11 i [ ∵ i = √-1 ]
So , By Quadratic Formula ,
x = -b ± √D / 2a
⇒ x = - ( -6 ) ± 2√11 i / 2 × 2
⇒ x = 6 ± 2√11 i/4 ⇒ x = 2 ( 3 ± √11 i ) 4
⇒ x = 3 ± √11 i / 2
Here. , Both the answers x ∈ R and x ∈ ϕ are correct . Let us understand this in two situations .
Situation 1 :-
x ∈ R , As the values of x are imaginary and we knows that all imaginary numbers are real . Hence , x ∈ R [ Not in the case of a Set ] .
Situation 2 :-
As we knows that. , imaginary numbers can't represented on a real axis where all elements of a given set ( except the empty set ) can be represented on a real axis. As the values of x can not be represented on a real axis . So. ,
x ∈ ϕ . Let us consider a set A which is representing the solution of the given modulus equation .
So , x ∈ ϕ if and only given that
A = { x | x is the solution of the modulus equation | -x² + 3x - 5 | + x² - 3x + 5 } = ϕ
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Hope it helps ya !