Question

What are the solutions of the systems? Y = x^2 + 3.9x + 4.4 y= 4.5x -2.8

Answer 1

Given equations:

$\quad \mathsf{y=x^{2}+3.9x+4.4}$ . . . (1)

$\quad \mathsf{y=4.5x-2.8}$ . . . (2)

To find: the solutions

Step-by-step explanation:

Using (2), from (1), we get

$\quad \mathsf{4.5x-2.8=x^{2}+3.9x+4.4}$

$\Rightarrow \mathsf{x^{2}+3.9x-4.5x+4.4+2.8=0}$

$\Rightarrow \mathsf{x^{2}-0.6x+7.2=0}$

Using quadratic formula, we get

$\quad \mathsf{x=\frac{-(-0.6)\pm \sqrt{(-0.6)^{2}-4\times 1\times 7.2}}{2\times 1}}$

$\Rightarrow \mathsf{x=\frac{0.6\pm\sqrt{0.36-28.8}}{2}}$

$\Rightarrow \mathsf{x=\frac{0.6\pm\sqrt{-28.44}}{2}}$

$\Rightarrow \mathsf{x=\frac{0.6\pm\sqrt{28.44}i}{2}}$ where $\mathsf{i=\sqrt{-1}}$

$\Rightarrow \mathsf{x=\frac{0.6\pm 6\sqrt{0.79}i}{2}}$

$\Rightarrow \mathsf{x=0.3\pm 3\sqrt{0.79}i}$

Now putting $\mathsf{x=0.3\pm 3\sqrt{0.79}i}$ in (2), we get

$\quad \mathsf{y=4.5\times (0.3\pm 3\sqrt{0.79}i)-2.8}$

$\Rightarrow \mathsf{y=1.35\pm 13.5\sqrt{0.79}i-2.8}$

$\Rightarrow \mathsf{y=-1.45\pm 13.5\sqrt{0.79}i}$

Answer:

Therefore the required solutions are

$\mathsf{x=0.3\pm 3\sqrt{0.79}i, y=-1.45\pm 13.5\sqrt{0.79}i}$