Solve the quadratic equation numerically (using tables of x- and y-values).

Mathematics

1 answer:

Stells [14]2 years ago

3 0

Answer:

Third Option: $x=6$ or $x=-3$

Step-by-step explanation:

$x^2+9x+18=0$

$(x+6)(x+3)=0$

$x=6$ or $x=-3$

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Sin^2(x/2)=sin^2x <br> Find the value of x

Veronika [31]

$\sin^2\dfrac x2=\sin^2x$
$\dfrac{1-\cos x}2=1-\cos^2x$
$1-\cos x=2-2\cos^2x$
$2\cos^2x-\cos x-1=0$
$(2\cos x+1)(\cos x-1)=0$
$\implies\begin{cases}2\cos x+1=0\\\cos x-1=0\end{cases}$ $\implies\begin{cases}\cos x=-\frac12\\\cos x=1\end{cases}$

The first case occurs in $0\le x$ for $x=\dfrac{2\pi}3$ and $x=\dfrac{4\pi}3$ . Extending the domain to account for all real $x$ , we have this happening for $x=\dfrac{2\pi}3+2n\pi$ and $\dfrac{4\pi}3+2n\pi$ , where $n\in\mathbb Z$ .

The second case occurs in $0\le x$ when $x=0$ , and extending to all reals we have $x=2n\pi$ for $n\in\mathbb Z$ , i.e. any even multiple of $\pi$ .