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Answers
Step-by-step explanation:
Given :-
5x - (3/2) = (5/2x) +14
To find :-
Solve the equation ?
Solution :-
Given equation is 5x - (3/2) = (5/2x) +14
=> [(5x×2)-3]/2 = [5+(14×2x)]/(2x)
=> (10x-3)/2 = (5+28x)/(2x)
On applying cross multiplication then
=> (2x)(10x-3) = (2)(5+28x)
=> (2x×10x)-(2x×3) = (2×5)+(2×28x)
=> 20x²-6x = 10+56x
=> 20x²-6x-10-56x = 0
=> 20x²-62x-10 = 0
=> 2(10x²-31x-5) = 0
=> 10x²-31x-5 = 0/2
=> 10x²-31x-5 = 0
On dividing by 10 both sides
=> (10x²/10)-(31x/10)-(5/10) = 0/10
=> x²-(31x/10)-(1/2) = 0
=> x²-(2/2)(31x/10) -(1/2) = 0
=> x²-2(31x/20) -(1/2) = 0
=> x²-2(31x/20) = 1/2
=> x²-2(x)(31/20) = 1/2
on adding (31/20)² both sides then
=> x²-2(x)(31/20) +(31/20)² = (1/2)+(31/20)²
LHS is in the form of a²-2ab+b²
Where, a = x and b = 31/20
We know that
(a-b)² = a²-2ab+b²
=> [x-(31/20)]² = (1/2)+(31/20)²
=> [x-(31/20)]² = (1/2)+(961/400)
=>[x-(31/20)]² = (200+961)/400
=>[x-(31/20)]² = 1161/400
=> x-(31/20) =±√(1161/400)
=> x -(31/20) = ±√1161 / 20
=> x = ±(√1161/20)+(31/20)
=> x = (31±√1161)/20
=> x = (31+√1161)/20 or (31-√1161)/20
Therefore, x = (31+√1161)/20 and
(31-√1161)/20
Answer :-
The solutions of the given equation are
(31+√1161)/20 and (31-√1161)/20
Check:-
If x = (31+√1161)/20 then
5x-(3/2) = (5/2x)+14
=> 5x-(5/2x) = 14+(3/2)
=> (10x²-5)/2x = (31/2)
=> 10x²-5 =31x
=> 10x²-31x-5 = 0
On taking LHS
10[31+√1161)/20]²-31(31+√1161)/20 -5
=> 10[(961+1161+62√1161)/400] -[(961+31√11161)/20] -5
=>[(2122+62√1161)/40]-[(961-31√1161-100)] /20
=> (2122+62√1161-1922-62√1161-200)/40
=> (200-200)/40
=> 0/40
=> 0
=>LHS = RHS is true for x values .
Used Method:-
- Completing the square method.
Solution:-
If a number satisfies the given equation (LHS = RHS) then it is called Root or Solution of the equation.