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Hunter-Best [27]
3 years ago
13

What is the solution to the equation

Mathematics
2 answers:
Dafna11 [192]3 years ago
7 0
\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
D:x\in\left< -\dfrac{5}{4};\ 6\right>

\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
Answer: t = -1.


Alborosie3 years ago
7 0

Answer:

-1

Step-by-step explanation:

:)

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in general, the graph of y = f(x-a) is a shift a units right if a is positive and shift of a units left if a is negative. for example, y = f(x-2) is 2 units right of f(x) and y=f(x+2) is 2 units LEFT.


in general, the graph of y = f(x) + a is a shift a units up if a is postiive and a units down if a is negtaive. For example, y = f(x) + 4 is 4 units up and y=f(x)-4 is a shift of 4 units down.

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