4x^2-28x+49=5 solve the quadratic by completing the square

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03-04-2014
Mathematics

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4x^2-28x+49=5 solve the quadratic by completing the square

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Lilith Lilith · Answer 1 · 2014-04-03T03:05:29+02:00

[tex]4x^2-28x+49=5 \\ \\4x^2-28x+49-5=0\\ \\4x^2-28x+44=0\\ \\ \Delta = b^{2}-4ac =(-28)^{2}-4*4*44=784 - 704 = 80 \\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{28- \sqrt{80}}{2*4}=\frac{-28-\sqrt{16*5}}{8}= \frac{-28-4\sqrt{ 5}}{8}=\\ \\=\frac{\not4^1(-7- \sqrt{ 5})}{\not8^2}=\frac{ -7- \sqrt{ 5} }{2}=[/tex]

[tex]x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{28+\sqrt{80}}{2*4}=\frac{-28+\sqrt{16*5}}{8}= \frac{-28+4\sqrt{ 5}}{8}=\\ \\=\frac{\not4^1(-7+ \sqrt{ 5})}{\not8^2}=\frac{ -7+ \sqrt{ 5} }{2} \\ \\Answer : x=\frac{ -7- \sqrt{ 5} }{2} \ \ or \ \ x =\frac{ -7+ \sqrt{ 5} }{2}[/tex]