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Running Head: MATHEMATICS: QUADRADIC EQUATIONMathematics: Quadradic EquationNameInstitution1MATHEMATICS: QUADRADIC EQUATION2If x+4y=40 and xy=6, what will be the value of x+2y? Please explain how youarrived at your answer.Assume that the value of x and y are two positive real numbers.X2+4y2= 40 (Eq i)xy= 6... (Eq ii)From (Eq ii) xy=6 divide both sides of the equation by y to get x.x=6/y. (Eq iii),Substitute Equation (iii) in Equation (i)(6/y)2 +4y2=40(36/y)2 +4y2=40 (Eq iv)Multiply both sides of Equation (iv) by y2.(36/y)2 x y2 +4y2 x y2 =40y236+4y4=40y2Collect like terms together4y4-40y2=-36Simplify 4y4-40y2=-36 by dividing both sides with 4 to get:Y4 40y2= -9.Y4-10y2 + 9=0 (Eq v)solve equation Eq (v)MATHEMATICS: QUADRADIC EQUATIONtake y=1 is (x+y)14 10(1) + 9=01-10+9=0Therefore, y2 -1 is a factor(Y-10y+9)/(y-1) = y-9Remember y-9 = (y+3) (y-3) when expandedAnd expanding y-1 gives us (y+1)(y-1)Thus, (y-10y+9)/(y-1)= (y+3)(y-3)(y+1)(y-1) (Tignol, 2015)Y=-3, or y=3 or y=-1 or y=1Substituting y=3 in equation (ii)Hence,xy=6 =3x=6x=6/3=2When y=3x=2Or when y=1x=6/1x=6Or when y= -3x=6/-33MATHEMATICS: QUADRADIC EQUATIONx=-2OrWhen y=-1x=6/-1x=-6Recall x and y are positive real numbers therefore y=3 or y=1When y=3, x=2 and when y=1, x=6.According to this.x + 4y = 40(x + 2y)2 = 40 + 4xyxy = 6x + 2y = 84MATHEMATICS: QUA ...


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