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IrinaK [193]
3 years ago
9

Answer true or false for 1-10 (picture above)

Mathematics
1 answer:
Softa [21]3 years ago
3 0

\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{a+b}\leq\sqrt{a}+\sqrt{b}\\\\\sqrt{a-b}\geq\sqrt{a}-\sqrt{b}\\----------------------\\1)\ \sqrt{72}=\sqrt{9\cdot8}=\sqrt{9}\cdot\sqrt{8}\qquad TRUE\\\\2)\ \sqrt{13-5}=\sqrt{13}-\sqrt5\qquad FALSE\\\\3)\ \sqrt{9\cdot10}=\sqrt9+\sqrt{10}\qquad FALSE\\\\4)\ \sqrt{18bx}=\sqrt{18}\cdot\sqrt{b}\cdot\sqrt{x}\qquad TRUE\ if\ b\geq0\ and\ x\geq0\\\\5)\ \sqrt{8\cdot12}=\sqrt8\cdot\sqrt{12}\qquad TRUE\\\\6)\ \sqrt{m+x}=\sqrt{m}+\sqrt{x}\qquad FALSE

7)\ \sqrt{100+64}=\sqrt{100}+\sqrt{64}\qquad FALSE\\\\8)\ \sqrt{30}=\sqrt{5\cdot6}=\sqrt5\cdot\sqrt6\qquad TRUE\\\\9)\ \sqrt{44}=\sqrt{22+22}=\sqrt{22}+\sqrt{22}\qquad FALSE\\\\10)\ \sqrt{10\cdot10}=\sqrt{10}\cdot\sqrt{10}\qquad TRUE

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Find all the zeros of the equation x^4-6x^2-7x-6=0
rusak2 [61]

Answer:

The zeros are

x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}

Step-by-step explanation:

We have been given the equation x^4-6x^2-7x-6=0

Use rational root theorem, we have

a_0=6,\:\quad a_n=1

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:3,\:6}{1}

-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2

=\left(x+2\right)\frac{x^4-6x^2-7x-6}{x+2}\\

=x^3-2x^2-2x-3

Again factor using the rational root test, we get

=\left(x+2\right)\left(x-3\right)\left(x^2+x+1\right)

Using the zero product rule, we have

x+2=0:\quad x=-2\\x-3=0:\quad x=3\\x^2+x+1=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}

Therefore, the zeros are

x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}


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Step-by-step explanation:

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