How do you solve #y = -2x^2 + 5x - 1#?

2 Answers
May 11, 2018

Answer:

The roots are:

#((5+sqrt17)/4,0)# and #((5-sqrt17)/4,0)#

The approximate roots are:

#(0.2192,0)# and #(2.281,0)#

Explanation:

Solve:

#y=-2x^2+5x-1# is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=-2#, #b=5#, #c=-1#

Substitute #0# for #y#.

#0=-2x^2+5x-1#

Solve using the quadratic equation.

#x=(-b+-sqrt(b^2-4ac))/(2*a)#

Plug in the known value.

#x=(-5+-sqrt(5^2-4*-2*-1))/(2*-2)#

Simplify.

#x=(-5+-sqrt(25-8))/(-4)#

#x=(-5+-sqrt17)/(-4)#

Simplify.

#x=(5+-sqrt17)/4#

#x=(5+sqrt17)/4,# #(5-sqrt17)/4#

The roots are:

#((5+sqrt17)/4,0)# and #((5-sqrt17)/4,0)#

The approximate roots are:

#(0.2192,0)# and #(2.281,0)#

graph{y=-2x^2+5x-1 [-10, 10, -5, 5]}

May 11, 2018

Answer:

#x = -0.22 " and " x = 2.28# for #y = 0#

Explanation:

When we say "solution" instead of the entire curve, we usually mean the place(s) were the function is zero. That is, the "roots" of the equation are at:

#-2x^2 + 5x - 1 = 0#

Now you can solve this be number of ways - factoring, quadratic formula, and graphing.

This one may be solved most quickly with the quadratic formula.

#x =( −b ± sqrt(b^2−4ac))/(2a)#
in this case, #a = -2, b = 5, c = -1#

#x =( −5 ± sqrt(5^2−4(-2)(-1)))/(2(-2))#

#x = (−5 ± sqrt(17))/(−4)#

#x = 5/4 ± sqrt(17)/4# ; #x = 5/4 ± 1.03#

#x = -0.22 " and " x = 2.28#