How do you solve #4x^4-5x^2-9=0#?
2 Answers
Explanation:
Set
Solve it with the usual quadratic formula to obtain
Now we must revert back to
Explanation:
#"make the substitution "u=x^2#
#rArr4u^2-5u-9=0#
#"using the a-c method to factor the quadratic"#
#"the factors of the product "4xx-9=-36#
#"which sum to - 5 are + 4 and - 9"#
#"split the middle term using these factors"#
#4u^2+4u-9u-9=0larrcolor(blue)"factor by grouping"#
#color(red)(4u)(u+1)color(red)(-9)(u+1)=0#
#"take out the "color(blue)" common factor "(u+1)#
#rArr(u+1)(color(red)(4u-9))=0#
#"equate each factor to zero and solve for u"#
#u+1=0rArru=-1#
#4u-9=0rArru=9/4#
#"change u back into terms of x"#
#rArrx^2=9/4rArrx=+-3/2larrcolor(red)"2 real roots"#
#x^2=-1rArrx=+-ilarrcolor(red)"2 complex roots"#