Which of the equations have at least two real roots?1) x4−5x2−36=02) x4−13x2+36=03) 4x4−10x2+25=0

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Kasen Alexander · Accepted Answer

Step 1This is a case of a &quot;hidden quadratic&quot; in that your equations are all really quadratic equations in a different variable. To see this, write u=x2. Then for example your first equation becomesu2−5u−36=0The temptation is then to look at the discriminant of this equation and conclude from there. But a solution of this equation gives us a value of u=x2, which will give two values of x if u&amp;gt;0, one (repeated) value if u=0 and no values if u&amp;lt;0. So in fact we have to solve this quadratic for u and go from there. In this case we haveu=5±25+1442=5±132=9,−4.This gives two values of u, but the case u=x2=−4 has no real solutions for x so we discard it. The case u=x2=9 has two solutions ±3 for x so overall there are two solutionsReally this is most of the way to solving the equations, although we only need to know the sign of the values of u, so for example we can see thatu=5±25+1442clearly gives one positive solution without explicitly calculating.

diesel817637dsf · Accepted Answer

Step 1x4−13x2+36=(x2+6)2−25x2=(x2+5x+6)(x2−5x+6)and both quadratic factors have real roots(positive discriminants).4x4−10x2+25=(2x2+5)2−30x2=(2x2+x30+5)(2x2−x30+5)and both quadratic factors have negative discriminants. OR4x4−10x2+25=(2x2−5)2+10x2is always positive

Which of the equations have at least two

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