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3 years ago

Which of the following are solutions to the equation below?

Mathematics

1 answer:

scoundrel [369]3 years ago

6 0

Answer:

The correct options for the solution values are:

$x=2\sqrt{2}-5$

$x=-2\sqrt{2}-5$

Step-by-step explanation:

Given the expression

$x^2+10x+25=8$

Subtract 25 from both sides

$x^2+10x+25-25=8-25$

Simplify

$x^2+10x=-17$

Add 25 or 5² to both sides

$x^2+10x+5^2=8$

$x^2+10x+5^2=\left(x+5\right)^2$

so the expression becomes

$\left(x+5\right)^2=8$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solve

$x+5=\sqrt{8}$

Subtract 5 from both sides

$x+5-5=2\sqrt{2}-5$

$x=2\sqrt{2}-5$

solve

$x+5=-\sqrt{8}$

Subtract 5 from both sides

$x+5-5=-2\sqrt{2}-5$

$x=-2\sqrt{2}-5$

Therefore, the solution to the equation

$x=2\sqrt{2}-5,\:x=-2\sqrt{2}-5$

Hence, the correct options for the solution values are:

$x=2\sqrt{2}-5$

$x=-2\sqrt{2}-5$

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3 years ago

Using substitution x+3y=3 3y-2x=12

frez [133]

Step-by-step explanation:

Given :-

x + 3y = 3
3y - 2x = 12

And we need to solve the equation using Substituting method . So on taking the first equation ,

$\rm\implies x + 3y = 3 \\\\\rm\implies x = 3 - 3y$

Put this value in (ii) :-

$\rm\implies 3y - 2x = 12 \\\\\rm\implies 3y - 2(3-3y)=12\\\\\rm\implies 3y -6-6y = 12 \\\\\rm\implies -3y = 18 \\\\\rm\implies y = -6$

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3 years ago

Read 2 more answers

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$