How do you find the zeros of the function #f(x)=(20x^2+11x-40)/(2x+5)#?
1 Answer
Jan 25, 2018
Explanation:
#"the denominator of f(x) cannot be zero as this would "#
#"make f(x) undefined"#
#"the zeros are found by equating the numerator to zero"#
#rArr"solve "20x^2+11x-40=0larrcolor(blue)"standard form"#
#"solve using the "color(blue)"quadratic formula"#
#"with "a=20,b=11" and "c=-40#
#•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)#
#rArrx=(-11+-sqrt(121+3200))/40#
#color(white)(rArrx)=(-11+-sqrt3321)/40=(-11+-9sqrt41)/40#
#rArrx=-11/40+-(9sqrt41)/40#
#rArrx=-11/40-(9sqrt41)/40" or "x=-11/40+(9sqrt41)/40#
#rArr"zeros are "x~~-1.72,x~~1.17" to 2 dec. places"#
