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IRINA_888 [86]
3 years ago
6

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

Mathematics
1 answer:
olga_2 [115]3 years ago
5 0

<em><u>The solution is (4, 4)</u></em>

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

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

<em><u>Substitute eqn 2 in eqn 1</u></em>

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

<em><u>Solve by quadratic equation</u></em>

\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

<em><u>Substitute x = 4 in eqn 2</u></em>

y = 2(4) - 4

y = 8 - 4

y = 4

Thus solution is (4, 4)

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