I need 4th and 8th ans by completing square method
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4)
2x² - 7x + 3 = 0
(√2x)² - 2(√2x)(7/2√2) + (7/2√2)² - (7/2√2)² +3 = 0
(√2x - 7/2√2)² - 49/8 + 3 = 0
(√2x - 7/2√2)² -49+24/8 = 0
(√2x - 7/2√2)² -25/8 = 0
(√2x - 7/2√2)² -(5/2√2)² = 0
(√2x - 7/2√2 + 5/2√2) (√2x - 7/2√2 - 5/2√2) = 0
(√2x - 7/2√2 + 5/2√2) = 0 (√2x - 7/2√2 - 5/2√2) = 0
(√2x -7+5/2√2) = 0 (√2x -7-5/2√2) = 0
(√2x -2/2√2) = 0 (√2x-12/2√2) = 0
(√2x -1/√2 ) = 0 (√2x-6/√2) = 0
x = 1/√2/√2 x = 6/√2/√2
x = 1/2 x = 6/2 = 3
4)
2x² + x - 4 = 0
(Take common 2 ) then ,
2(x² + x/2 - 4 ) = 0
x² + x/2 - 4 = 0/2
x² + x/2 - 4 = 0
(x)² + 2(x)(1/4) + (1/4)² - (1/4)² - 4 = 0
(x + 1/4)² - 1/16 - 4 = 0
(x + 1/4)² -1-64/16 = 0
(x + 1/4)² -65/16 = 0
(x + 1/4)² -(√65/4)² = 0
(x + 1/4 + √65/4) (x + 1/4 - √65/4) = 0
(x + 1/4 + √65/4) = 0 (x + 1/4 - √65/4) = 0
(x + 1+√65/4) = 0 (x + 1-√65/4) = 0
x = -1-√65/4 x = -1+√65/4
2x² - 7x + 3 = 0
(√2x)² - 2(√2x)(7/2√2) + (7/2√2)² - (7/2√2)² +3 = 0
(√2x - 7/2√2)² - 49/8 + 3 = 0
(√2x - 7/2√2)² -49+24/8 = 0
(√2x - 7/2√2)² -25/8 = 0
(√2x - 7/2√2)² -(5/2√2)² = 0
(√2x - 7/2√2 + 5/2√2) (√2x - 7/2√2 - 5/2√2) = 0
(√2x - 7/2√2 + 5/2√2) = 0 (√2x - 7/2√2 - 5/2√2) = 0
(√2x -7+5/2√2) = 0 (√2x -7-5/2√2) = 0
(√2x -2/2√2) = 0 (√2x-12/2√2) = 0
(√2x -1/√2 ) = 0 (√2x-6/√2) = 0
x = 1/√2/√2 x = 6/√2/√2
x = 1/2 x = 6/2 = 3
4)
2x² + x - 4 = 0
(Take common 2 ) then ,
2(x² + x/2 - 4 ) = 0
x² + x/2 - 4 = 0/2
x² + x/2 - 4 = 0
(x)² + 2(x)(1/4) + (1/4)² - (1/4)² - 4 = 0
(x + 1/4)² - 1/16 - 4 = 0
(x + 1/4)² -1-64/16 = 0
(x + 1/4)² -65/16 = 0
(x + 1/4)² -(√65/4)² = 0
(x + 1/4 + √65/4) (x + 1/4 - √65/4) = 0
(x + 1/4 + √65/4) = 0 (x + 1/4 - √65/4) = 0
(x + 1+√65/4) = 0 (x + 1-√65/4) = 0
x = -1-√65/4 x = -1+√65/4
sowmiya7:
u have answered 2nd q I asked 4th and 8th
sorry for that mistake now it is corrected
ya thanks
but we're is the 4th ans
see once again the answer
where
But ans is wrong
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get your answer by completing square method
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I need 4th ans
plzz
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