PLEASE HELP it hasn’t been a good day and i don’t feel like doing this

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

4 0

Answer:

Both are quadrant

Step-by-step explanation:

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X2 - 6x + 12 and y = 2x - 4, algebraically are

olga_2 [115]

The solution is (4, 4)

Solution:

Given system of equations are:

$y = x^2 - 6x + 12 ------ eqn\ 1\\\\y = 2x - 4 ---------- eqn\ 2$

Substitute eqn 2 in eqn 1

$x^2 - 6x + 12 = 2x - 4$

Make the right side of equation 0

$x^2 - 6x + 12 - 2x + 4 = 0\\\\x^2 -8x + 16 = 0$

Solve by quadratic equation

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=-8,\:c=16:\\\\x=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:16}}{2\cdot \:1}\\\\x=\frac{-\left(-8\right)\pm \sqrt{0}}{2\cdot \:1}\\\\x = \frac{8}{2}\\\\x = 4$