Answer: [tex]x^{2} +1 = 3x[/tex] [tex]x^{2} – 3x + 1=0[/tex] For a quadratic equation of the form ax²+bx+c=0 the solutions are [tex]x = \frac{-b +- \sqrt{b^{2} -4ac} }{2a}[/tex] for a=1, b= -3, c=1 [tex]x = \frac{-(-3)+- \sqrt{(-3)^{2} -4(1)(1)} }{2(1)}[/tex] [tex]x = \frac{-(-3)+-\sqrt{5} }{2}[/tex] [tex]x = \frac{3+\sqrt{5} }{2} , \frac{3-\sqrt{5} }{2}[/tex] Reply
x²+1 = 3x x²-3x +1 = 0 x² -3x +1 a = 1 b = -3 c = 1 applying on Quadratic equation, x = -(-3) +/-√[(-3)² – (4×1×1)] / 2 = 3 +/- √[ 9 – 4 ] / 2 = 3 +/- √5 / 2 x = 3 + √5 /2 x = 3 – √5 / 2 = 3 + 2.23 / 2 = 3 – 2.23 / 2 = 2.615 = 0.385 Reply
Answer:
[tex]x^{2} +1 = 3x[/tex]
[tex]x^{2} – 3x + 1=0[/tex]
For a quadratic equation of the form ax²+bx+c=0 the solutions are
[tex]x = \frac{-b +- \sqrt{b^{2} -4ac} }{2a}[/tex]
for a=1, b= -3, c=1
[tex]x = \frac{-(-3)+- \sqrt{(-3)^{2} -4(1)(1)} }{2(1)}[/tex]
[tex]x = \frac{-(-3)+-\sqrt{5} }{2}[/tex]
[tex]x = \frac{3+\sqrt{5} }{2} , \frac{3-\sqrt{5} }{2}[/tex]
x²+1 = 3x
x²-3x +1 = 0
x² -3x +1 a = 1 b = -3 c = 1
applying on Quadratic equation,
x = -(-3) +/-√[(-3)² – (4×1×1)] / 2
= 3 +/- √[ 9 – 4 ] / 2
= 3 +/- √5 / 2
x = 3 + √5 /2 x = 3 – √5 / 2
= 3 + 2.23 / 2 = 3 – 2.23 / 2
= 2.615 = 0.385