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Evaluating expression with Integer part and Fraction part of a nested radical
Dealing with negatives in radicalCubic polynomial - radical expression of rootsSolving the 4th Degree Equation $x^4- 8sqrt3x^2 - 16 = 0$Calculate radical expression without a calculatorFind another polynomial with the same roots as a polynomial with radical coefficientsSimplifying an expression with a nested radicalHow do I compute the limit of a radical outside of a fraction?Radical additive expression in relation to partial derivative.Simplifying nested radicals with higher-order radicalsHow to solve this radical expression
$begingroup$
Let
$$A= sqrt93+28sqrt11$$
if $B$ is the integer part of $A$ and $C$ is the fraction part of $C$, what is the value of
$$B+C^2$$
I tried manipulating it by setting
$$ A=B+C$$
but I can't transform it into the expression, do I need approximate the integer part of A in order to solve this or is there another way?
radicals
edited 2 days ago
Maria Mazur
48.3k1260121
asked 2 days ago
SuperMage1SuperMage1
947211
$endgroup$
add a comment |
$begingroup$
Let
$$A= sqrt93+28sqrt11$$
if $B$ is the integer part of $A$ and $C$ is the fraction part of $C$, what is the value of
$$B+C^2$$
I tried manipulating it by setting
$$ A=B+C$$
but I can't transform it into the expression, do I need approximate the integer part of A in order to solve this or is there another way?
radicals
edited 2 days ago
Maria Mazur
48.3k1260121
asked 2 days ago
SuperMage1SuperMage1
947211
$endgroup$
$begingroup$
Presumably you mean $C$ is the fractional part of $A$
$endgroup$
– Ross Millikan
2 days ago
$begingroup$
What do you mean by fractional part? That number is irrational.
$endgroup$
– Allawonder
2 days ago
add a comment |
$begingroup$
Let
$$A= sqrt93+28sqrt11$$
if $B$ is the integer part of $A$ and $C$ is the fraction part of $C$, what is the value of
$$B+C^2$$
I tried manipulating it by setting
$$ A=B+C$$
but I can't transform it into the expression, do I need approximate the integer part of A in order to solve this or is there another way?
radicals
edited 2 days ago
Maria Mazur
48.3k1260121
asked 2 days ago
SuperMage1SuperMage1
947211
$endgroup$
Let
$$A= sqrt93+28sqrt11$$
if $B$ is the integer part of $A$ and $C$ is the fraction part of $C$, what is the value of
$$B+C^2$$
I tried manipulating it by setting
$$ A=B+C$$
but I can't transform it into the expression, do I need approximate the integer part of A in order to solve this or is there another way?
radicals
radicals
edited 2 days ago
Maria Mazur
48.3k1260121
asked 2 days ago
SuperMage1SuperMage1
947211
edited 2 days ago
Maria Mazur
48.3k1260121
asked 2 days ago
SuperMage1SuperMage1
947211
edited 2 days ago
Maria Mazur
48.3k1260121
edited 2 days ago
Maria Mazur
48.3k1260121
edited 2 days ago
Maria Mazur
48.3k1260121
48.3k1260121
asked 2 days ago
SuperMage1SuperMage1
947211
asked 2 days ago
SuperMage1SuperMage1
947211
asked 2 days ago
SuperMage1SuperMage1
947211
947211
$begingroup$
Presumably you mean $C$ is the fractional part of $A$
$endgroup$
– Ross Millikan
2 days ago
$begingroup$
What do you mean by fractional part? That number is irrational.
$endgroup$
– Allawonder
2 days ago
add a comment |
$begingroup$
Presumably you mean $C$ is the fractional part of $A$
$endgroup$
– Ross Millikan
2 days ago
$begingroup$
What do you mean by fractional part? That number is irrational.
$endgroup$
– Allawonder
2 days ago
$begingroup$
Presumably you mean $C$ is the fractional part of $A$
$endgroup$
– Ross Millikan
2 days ago
$begingroup$
Presumably you mean $C$ is the fractional part of $A$
$endgroup$
– Ross Millikan
2 days ago
$begingroup$
What do you mean by fractional part? That number is irrational.
$endgroup$
– Allawonder
2 days ago
$begingroup$
What do you mean by fractional part? That number is irrational.
$endgroup$
– Allawonder
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's find intgers $x,y$ such that $$(x+ysqrt11)^2= 93+28sqrt11$$
Then $$x^2+11y^2 =93 ;;;wedge ;;; 2xy = 28$$
Since $11y^2<99implies y^2 <9 implies |y|leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$boxedsqrt93+28sqrt11 = 7+2sqrt11$$
Since $$13= 7+2cdot 3<7+2sqrt11 <7+2cdot sqrt49over 4 = 14$$
we see $B =13$...
answered 2 days ago
Maria MazurMaria Mazur
48.3k1260121
$endgroup$
add a comment |
$begingroup$
Hint:
$(7+2sqrt11)^2=cdots$
answered 2 days ago
drhabdrhab
103k545136
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's find intgers $x,y$ such that $$(x+ysqrt11)^2= 93+28sqrt11$$
Then $$x^2+11y^2 =93 ;;;wedge ;;; 2xy = 28$$
Since $11y^2<99implies y^2 <9 implies |y|leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$boxedsqrt93+28sqrt11 = 7+2sqrt11$$
Since $$13= 7+2cdot 3<7+2sqrt11 <7+2cdot sqrt49over 4 = 14$$
we see $B =13$...
answered 2 days ago
Maria MazurMaria Mazur
48.3k1260121
$endgroup$
add a comment |
$begingroup$
Let's find intgers $x,y$ such that $$(x+ysqrt11)^2= 93+28sqrt11$$
Then $$x^2+11y^2 =93 ;;;wedge ;;; 2xy = 28$$
Since $11y^2<99implies y^2 <9 implies |y|leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$boxedsqrt93+28sqrt11 = 7+2sqrt11$$
Since $$13= 7+2cdot 3<7+2sqrt11 <7+2cdot sqrt49over 4 = 14$$
we see $B =13$...
answered 2 days ago
Maria MazurMaria Mazur
48.3k1260121
$endgroup$
add a comment |
$begingroup$
Let's find intgers $x,y$ such that $$(x+ysqrt11)^2= 93+28sqrt11$$
Then $$x^2+11y^2 =93 ;;;wedge ;;; 2xy = 28$$
Since $11y^2<99implies y^2 <9 implies |y|leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$boxedsqrt93+28sqrt11 = 7+2sqrt11$$
Since $$13= 7+2cdot 3<7+2sqrt11 <7+2cdot sqrt49over 4 = 14$$
we see $B =13$...
answered 2 days ago
Maria MazurMaria Mazur
48.3k1260121
$endgroup$
Let's find intgers $x,y$ such that $$(x+ysqrt11)^2= 93+28sqrt11$$
Then $$x^2+11y^2 =93 ;;;wedge ;;; 2xy = 28$$
Since $11y^2<99implies y^2 <9 implies |y|leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$boxedsqrt93+28sqrt11 = 7+2sqrt11$$
Since $$13= 7+2cdot 3<7+2sqrt11 <7+2cdot sqrt49over 4 = 14$$
we see $B =13$...
answered 2 days ago
Maria MazurMaria Mazur
48.3k1260121
answered 2 days ago
Maria MazurMaria Mazur
48.3k1260121
answered 2 days ago
Maria MazurMaria Mazur
48.3k1260121
answered 2 days ago
Maria MazurMaria Mazur
48.3k1260121
48.3k1260121
add a comment |
add a comment |
$begingroup$
Hint:
$(7+2sqrt11)^2=cdots$
answered 2 days ago
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Hint:
$(7+2sqrt11)^2=cdots$
answered 2 days ago
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Hint:
$(7+2sqrt11)^2=cdots$
answered 2 days ago
drhabdrhab
103k545136
$endgroup$
Hint:
$(7+2sqrt11)^2=cdots$
answered 2 days ago
drhabdrhab
103k545136
answered 2 days ago
drhabdrhab
103k545136
answered 2 days ago
drhabdrhab
103k545136
answered 2 days ago
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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-radicals
$begingroup$
Presumably you mean $C$ is the fractional part of $A$
$endgroup$
– Ross Millikan
2 days ago
$begingroup$
What do you mean by fractional part? That number is irrational.
$endgroup$
– Allawonder
2 days ago