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3 years ago

What are the possible numbers of positive, negative, and complex zeros of f(x)=x^6-x^5-x^4+4x^3-12x^2+12?

1 answer:

8 0

$f(x)=x^6-x^5-x^4+4x^3-12x^2+12\\\\12:\{\pm1;\ \pm2;\ \pm3;\ \pm4;\ \pm6;\ \pm12\}\\1:\{\pm1\}\\\\Answer:\boxed{\{\pm1;\ \pm2;\ \pm3;\ \pm4;\ \pm6;\ \pm12\}}$

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On a trip of d miles to another city, a truck driver's average speed was x miles per hour. On the return trip, the average speed

max2010maxim [7]

Mile x was 15 miles long per hour while Y was 15 miles per hour on the two trips

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3 years ago

What are the solutions of x^2-2x+5=0

sweet [91]

The solution of $x^{2}-2 x+5=0$ are 1 + 2i and 1 – 2i

Solution:

Given, equation is $x^{2}-2 x+5=0$

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ --- (1)

Let us determine the nature of roots:

Here in $x^{2}-2 x+5=0$ a = 1 ; b = -2 ; c = 5

$b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$

Since $b^2 - 4ac < 0$ , the roots obtained will be complex conjugates.

Now plug in values in eqn 1, we get,

$x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}$

On solving we get,

$x=\frac{2 \pm \sqrt{4-20}}{2}$

$x=\frac{2 \pm \sqrt{-16}}{2}$

$x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}$

we know that square root of -1 is "i" which is a complex number

$\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}$

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

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3 years ago

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3 years ago

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3 years ago