Question

What are the zeros of the quadratic function f(x) = 6x2 – 24x + 1? fast please

Answer 1

Answer:      6(x²-4x+4-4)+1=0, 6(x-2)²-24+1=0, 6(x-2)²=23, x-2=±√(23/6), x=2±√(23/6)=2±1.95789, so x=3.95789 or 0.04211 approx. these are the zeros.

step by step explanation:

\boxed{\boxed{\dfrac{12+\sqrt{138}}{6},\ \dfrac{12-\sqrt{138}}{6}}}

Solution-

The quadratic function is,

6x^2-24x + 1

a = 6, b = -24, c = 1

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

=\dfrac{-(-24)\pm \sqrt{-24^2-4\cdot 6\cdot 1}}{2\cdot 6}

=\dfrac{24\pm \sqrt{576-24}}{12}

=\dfrac{24\pm \sqrt{552}}{12}

=\dfrac{24\pm 2\sqrt{138}}{12}

=\dfrac{12\pm \sqrt{138}}{6}

=\dfrac{12+\sqrt{138}}{6},\ \dfrac{12-\sqrt{138}}{6}