Consider the quadratic equation x2=4x-5. How many solutions does the equation have

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Consider the quadratic equation x2=4x-5. How many solutions does the equation have

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    2021-09-09T16:22:38+00:00

    \bf x^2=4x-5\implies x^2-4x+5=0 \\\\\\ \begin{array}{lcccll} & 1 x^2& -4 x& +5&=0\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \\\\\\ discriminant\implies b^2-4ac= \begin{cases} 0&\textit{one solution}\\ positive&\textit{two solutions}\\ negative&\textit{no solution} \end{cases}

    now, plugging in those values for the discriminant, we’d get

    (-4)² – 4(1)(5).

    0
    2021-09-09T16:22:39+00:00

    the quadratic formula is =(-b+-sqrt(b^2-4ac))/2a
    as you notice the term under the square root is b^2-4ac if it is postive then the equation clearly will have two real soultions if it is negative then the equation will have two imaginary soultion if it is zero then the the equation will have one soultion
    so let us calculate b^2-4ac for our given equation
    x^2=4x-5 so let us write it in general form which is ax^2+bx+c=0
    subtracting 4x from both sides
    x^2-4x=-5
    adding 5 to both sides
    x^2-4x+5=0
    a=1,b=-4,c=5
    b^2-4ac=(-4)^2-4(1)(5)=16-20=-4
    which means the equation has two imaginary soultions

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