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Who can solve quadratic and simultaneous quations ?

April 26, 2017answer knowledge quadratic simultaneous Yedda

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Who can solve quadratic and simultaneous quations ?

2 Answers

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1) (2x+3)sq=4 => (2x+3)sq=(2)sq =>2x+3=2 and 2x+3=-2 x=-1/2 ans x=-5/2 ANS 2),,, x – 2y = 7 ——————–1 x squared +4ysquared= 37 ————————2 putting x=7+2y in 2 equationm we get (7+2y)sq+4ysq=37 by solving we get ysq+7y+3=0 solving using quadratic formula we get y= (-7 +-(sqrt37))/2 ANS

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The parabola y = (2x+3)^2 crosses the horizontal line y = 4, in two places which are when (2x+3)^2 = 4 so this is the interpretation of this equation on a graph. We will get the two results that we expect, because 4 has two square roots, +2 and -2 Taking the square roots of both sides. Either 2x+3 = 2 or 2x+3 = -2 x = -1/2 or x = -5/2 2) Once again we may interpret this as the line x – 2y = 7 crossing the parabola x^2 +4y^2 = 37 which will be in two places. As we expect two results we substitute from the line equation into the parabola equation, (using either x or y) Substituting x = 2y + 7 yields (2y + 7)^2 + 4y^2 – 37 = 0 8y^2 + 28y +12 = 0 dividing this by four, 2y^2 + 7y +3 = 0 [if you do not see how to factor at this stage use the quadratic formula].