 Solve the equation: $$x^4-16x^2-225=0$$
Explanation
Substitute $$y = x^2$$: $$y^2-16y-225=0$$ Find discriminant of quadratic equation: $$D=(-16)^2-4\cdot1\cdot(-225)=256+900=1156$$ Discriminant is greater than zero, so there are two (unequal) real solutions of the equation: $$y_{1,2}=\dfrac{16\pm34}{2\cdot1}={25,-9}$$ There two quadratic equations: $$(1) x^2=-9$$ $$(2) x^2=25$$ Quadratic equation (1) has no real solutions. Quadratic equation (2) has two (unequal) real solutions: $$x_{1,2}=\pm5$$
Theory
• Solution of quadratic equations. A quadratic equation is any equation having the form $$ax^2+bx+c=0$$ where $$x$$ represents an unknown, and $$a,b$$ and $$c$$ represent known numbers such that $$a$$ is not equal to 0. There are two stages in solving quadratic equations. 1) Find discriminant using formula $$D=b^2-4ac$$
2) Check the sign of discriminant: 2.1) If $$D>0$$, then equation has two (unequal) real solutions $$x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2\cdot a}$$
2.2) If $$D=0$$, then equation has two (equal) real solutions $$x=\dfrac{-b}{2\cdot a}$$
2.3) If $$D<0$$, then equation has no real solutions.

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