solve the equation 2x^-5x+3=0 by the method of completing the square
Answers
Solution:
2x²-5x+3 = 0
=> 2x²-5x = -3
Divide each term by 2 , we get
=> x²-(5x/2) = -3/2
=> x² -2*x*(5/4) = -3/2
=> x²-2*x*(5/4)+(5/4)²=-3/2+(5/4)²
=> (x-5/4)² = -3/2+25/16
=> (x-5/4)²=(-24+25)/16
=> (x-5/4)² = 1/16
=> x-5/4 = ±√(1/16)
=> x = 5/4± 1/4
=> x = (5±1)/4
=> x = (5+1)/4 or x = (5-1)/2
=> x = 3/2 or x = 2
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Answer:
Step 1: Divide by the coefficient of x²
Coefficient of x² is 2. Hence dividing the whole equation by 2 we get,
⇒ x² - 5/2 x + 3/2 = 0
Step 2: Transpose the constant terms to the RHS
⇒ x² - 5/2 x = -3/2
Step 3: Divide the coefficient of x by 2 and add the result's square on both sides.
⇒ Coefficient of x = -5/2
Dividing by 2 we get,
⇒ -5/2 ÷ 2 = -5/4
Adding it on both sides we get,
⇒ x² - 5/2 x + (5/4)² = -3/2 + (5/4)²
Step 4: Write LHS in the form of ( a + b )²
⇒ ( x - 5/4 )² = -3/2 + 25/16
Taking LCM on the RHS, we get,
⇒ ( x - 5/4 )² = ( -24 + 25 ) / 16
⇒ ( x - 5/4 )² = 1/16
Taking Square root on both sides, we get,
⇒ x - 5/4 = ± 1/4
⇒ x = 5/4 ± 1/4
∴ x = 5/4 + 1/4 and 5/4 - 1/4
⇒ x = 6/4 and 4/4
⇒ x = 3/2 and 1
Hence the zeros or the roots of the equation are 2/3 and 1.
Hope it helped !!