How do you solve #4x^4-5x^2-9=0#?

2 Answers
May 14, 2018

Answer:

#x = \pm 3/2#

Explanation:

Set #t=x^2#. Then, #t^2 = (x^2)^2 = x^4#. The equation becomes

#4t^2-5t-9=0#

Solve it with the usual quadratic formula to obtain #t_1 = -1# and #t_2 = 9/4#

Now we must revert back to #x#:

#t_1 = -1 \implies x^2 = -1#, which is impossible.

#t_2 = 9/4 \implies x^2 = 9/4 \implies x = pm 3/2#

May 14, 2018

Answer:

#x=+-3/2" or "x=+-i#

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"#