Mathematics

Question

Determine the type and number of solutions of 4x^2-5x+1=0.
A. Two real solutions
B. One real solution
C. Two imaginary solutions

1 Answer

  • 4x² - 5x + 1 = 0
    a = 4; b = - 5, c = 1
    Δ = b² - 4.a.c
    Δ = (-5)² - 4.4.1
    Δ = 25 - 16
    Δ = 9

    x = - b 
    ± √Δ / 2.a
    [tex]x = \frac{-5\± \sqrt{9} }{2*4} [/tex]
    [tex]x = \frac{-5\±3}{8} [/tex]
    [tex]x' = \frac{-5-3}{8} = \frac{-8}{8} = -1[/tex]
    [tex]x' = \frac{-5+3}{8} = \frac{-2}{8} (\div2) = \frac{-1}{4}[/tex]


    A. Two real solutions (x' = -1 and x'' = -1/4)
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