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antiseptic1488 [7]
4 years ago
7

Solve the system y = -x + 7 and y = -0.5(x - 3)^2 + 8

Mathematics
1 answer:
baherus [9]4 years ago
4 0
Both equal y so set equal to each other
-x+7=-0.5(x-3)^2+8
times both sides by -2
2x-14=(x-3)^2-16
expand
2x-14=x^2-6x+9-16
add 14-2x to both sides
0=x^2-8x+7
factor
0=(x-7)(x-1)
set to zero
0=x-7
7=x

0=x-1
1=x

so find y

y=-x+7
y=-1+7
y=6
or

y=-7+7
y=0

so the solutions are
(-1,6) and (-7,0)
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\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6

therefore
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\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs
9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D
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7 0
3 years ago
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