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 Quadratic Equations Equations in Quadratic Form Some equations are not quadratic equations, but are in the same form. To solve equations such as that, make a substitution, solve for the new variable, and then solve for the original variable $\fs2x^4-5x^2+4\;=\;0\;\Rightarrow\;\;y=x^2$$\fs2y^2-5y+4\;=\;0\;\Rightarrow\;y=\frac{5\pm\sqrt{(-5)^2-4\;1\;4}}{2\;1}\;\Rightarrow\;y=\frac{5\pm\sqrt{25-16}}{2}\;\Rightarrow\;y=\frac{5\pm3}{2}$$\fs2y_1=\frac{5+3}{2}=\frac{8}{2}=4\;\;\;\;y_2=\frac{5-3}{2}=\frac{1}{2}=1$These give$\fs2x^2\;=\;y_1\;\Rightarrow\;\;x^2\;=\;4\;\Rightarrow\;\;x\;=\;\pm2$ $x^2\;=\;y_2\;\Rightarrow\;x^2\;=\;1\;\Rightarrow\;\;x\;=\;\pm1\;\;\;$ Solve $\fs1x^4-15x^2-16\;=\;0$ No solution One solution x= Two solutions x1= x2= Three solutions x1= x2= x3= Four solutions x1= x2= x3= x4=