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Find y in this equation

How find this equationIs it possible to solve this equation by hand?Is this a quadratic equation?How to solve the equation $exp(3t) - exp(2t) = A$How do I solve this equation :$barz-iz²=-sqrt3-3i$ without using identity way?How do you factor this equation that has square roots involved?What is the name for an equation that has both an exponential variable and a variable in the base?Solving simple equations for two variables

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3

1

$begingroup$

I would like to solve this equation for y. Any tips? It seems like you cant really do it analytically?

$$sqrt4+(y-6)^2+sqrt16+(y-3)^2=3cdotsqrt5$$

algebra-precalculus

share|cite|improve this question

edited 6 hours ago

Aqua

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Allawonder where are you getting a quartic?
  $endgroup$
  – Simply Beautiful Art
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @SimplyBeautifulArt My bad. My judgement was quite premature. I later saw that terms higher than the square vanished, as explained in my answer below.
  $endgroup$
  – Allawonder
  7 hours ago
  

  
  
  
  

add a comment | 

3

1

$begingroup$

I would like to solve this equation for y. Any tips? It seems like you cant really do it analytically?

$$sqrt4+(y-6)^2+sqrt16+(y-3)^2=3cdotsqrt5$$

algebra-precalculus

share|cite|improve this question

edited 6 hours ago

Aqua

56.5k1515 gold badges7171 silver badges139139 bronze badges

asked 8 hours ago

user2276094user2276094

1611 bronze badge

New contributor

user2276094 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Allawonder where are you getting a quartic?
  $endgroup$
  – Simply Beautiful Art
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @SimplyBeautifulArt My bad. My judgement was quite premature. I later saw that terms higher than the square vanished, as explained in my answer below.
  $endgroup$
  – Allawonder
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I would like to solve this equation for y. Any tips? It seems like you cant really do it analytically?

$$sqrt4+(y-6)^2+sqrt16+(y-3)^2=3cdotsqrt5$$

algebra-precalculus

share|cite|improve this question

edited 6 hours ago

Aqua

56.5k1515 gold badges7171 silver badges139139 bronze badges

asked 8 hours ago

user2276094user2276094

1611 bronze badge

New contributor

user2276094 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 6 hours ago

Aqua

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asked 8 hours ago

user2276094user2276094

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share|cite|improve this question

edited 6 hours ago

Aqua

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user2276094user2276094

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edited 6 hours ago

Aqua

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edited 6 hours ago

Aqua

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asked 8 hours ago

user2276094user2276094

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asked 8 hours ago

user2276094user2276094

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4 Answers
4

active

oldest

votes

3

$begingroup$

Set $y-3=z,$ then square both sides and simplify to get $$z^2-3z-8=-sqrt4+(z-3)^2sqrt16+z^2.$$ Squaring again and simplifying the right hand side gives $$(z^2-3z-8)^2=(z^2-6z+13)(16+z^2).$$ Now let $16+z^2=w,$ then expand and simplify to get $$z^2+16z+64=5w.$$ Substituting back for $w,$ simplifying and factoring gives the quadratic $$(z-2)^2=0.$$ Thus, $z=y-3=2,$ so that if there is a solution, then it has to be $y=5.$ Substituting this in the original equation assures us that this is indeed the only solution.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

AllawonderAllawonder

3,74788 silver badges1818 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What does suggest $sqrtdfrac4+(y-6)^25+sqrtdfrac16+(y-3)^25=3$? It is clear that $y=5$ is a root.
  $endgroup$
  – Piquito
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Consider points $A(2,6)$, $B(-4,3)$ and a point $P(0,y)$ on y-axis. Then notice that given equation is actually $$AP+BP= AB$$ so $P$ is (by triangle inequality ) on a line $AB$ so it is an intersection point of the line $AB$: $$ y= 1over 2x+5$$ and $y$-axis, so $y=5$.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

AquaAqua

56.5k1515 gold badges7171 silver badges139139 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

Hint: After squaring one times we get
$$2sqrt4+(y-6)^2sqrt16+(y-3)^2=25-(y-6)^2-(y-3)^2$$
squaring again and simplfying we get
$$4 left(46 y^2-450 y+975right)=0$$

share|cite|improve this answer

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

86.6k44 gold badges2929 silver badges7171 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

You can factor out $2$ at the second summand.

$$sqrtunderbrace4+(y-6)^2_=5+2cdot sqrtunderbrace4+left(fracy-32right)^2_=5=sqrt5+2cdotsqrt5$$

Now we see that the following equations has to be true at the same time.

1. $(y-6)^2=1 Rightarrow y_1=7,y_2=5$




2. $left(fracy-32right)^2=1Rightarrow y_1=5,y_2=1$

Thus the solution is $y=...$

share|cite|improve this answer

answered 7 hours ago

callculuscallculus

20k33 gold badges1717 silver badges3232 bronze badges

$endgroup$

add a comment | 

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3

$begingroup$

Set $y-3=z,$ then square both sides and simplify to get $$z^2-3z-8=-sqrt4+(z-3)^2sqrt16+z^2.$$ Squaring again and simplifying the right hand side gives $$(z^2-3z-8)^2=(z^2-6z+13)(16+z^2).$$ Now let $16+z^2=w,$ then expand and simplify to get $$z^2+16z+64=5w.$$ Substituting back for $w,$ simplifying and factoring gives the quadratic $$(z-2)^2=0.$$ Thus, $z=y-3=2,$ so that if there is a solution, then it has to be $y=5.$ Substituting this in the original equation assures us that this is indeed the only solution.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

AllawonderAllawonder

3,74788 silver badges1818 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What does suggest $sqrtdfrac4+(y-6)^25+sqrtdfrac16+(y-3)^25=3$? It is clear that $y=5$ is a root.
  $endgroup$
  – Piquito
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Set $y-3=z,$ then square both sides and simplify to get $$z^2-3z-8=-sqrt4+(z-3)^2sqrt16+z^2.$$ Squaring again and simplifying the right hand side gives $$(z^2-3z-8)^2=(z^2-6z+13)(16+z^2).$$ Now let $16+z^2=w,$ then expand and simplify to get $$z^2+16z+64=5w.$$ Substituting back for $w,$ simplifying and factoring gives the quadratic $$(z-2)^2=0.$$ Thus, $z=y-3=2,$ so that if there is a solution, then it has to be $y=5.$ Substituting this in the original equation assures us that this is indeed the only solution.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

AllawonderAllawonder

3,74788 silver badges1818 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What does suggest $sqrtdfrac4+(y-6)^25+sqrtdfrac16+(y-3)^25=3$? It is clear that $y=5$ is a root.
  $endgroup$
  – Piquito
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Set $y-3=z,$ then square both sides and simplify to get $$z^2-3z-8=-sqrt4+(z-3)^2sqrt16+z^2.$$ Squaring again and simplifying the right hand side gives $$(z^2-3z-8)^2=(z^2-6z+13)(16+z^2).$$ Now let $16+z^2=w,$ then expand and simplify to get $$z^2+16z+64=5w.$$ Substituting back for $w,$ simplifying and factoring gives the quadratic $$(z-2)^2=0.$$ Thus, $z=y-3=2,$ so that if there is a solution, then it has to be $y=5.$ Substituting this in the original equation assures us that this is indeed the only solution.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

AllawonderAllawonder

3,74788 silver badges1818 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

AllawonderAllawonder

3,74788 silver badges1818 bronze badges

answered 8 hours ago

AllawonderAllawonder

3,74788 silver badges1818 bronze badges

answered 8 hours ago

AllawonderAllawonder

3,74788 silver badges1818 bronze badges

1

$begingroup$

Consider points $A(2,6)$, $B(-4,3)$ and a point $P(0,y)$ on y-axis. Then notice that given equation is actually $$AP+BP= AB$$ so $P$ is (by triangle inequality ) on a line $AB$ so it is an intersection point of the line $AB$: $$ y= 1over 2x+5$$ and $y$-axis, so $y=5$.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

AquaAqua

56.5k1515 gold badges7171 silver badges139139 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Consider points $A(2,6)$, $B(-4,3)$ and a point $P(0,y)$ on y-axis. Then notice that given equation is actually $$AP+BP= AB$$ so $P$ is (by triangle inequality ) on a line $AB$ so it is an intersection point of the line $AB$: $$ y= 1over 2x+5$$ and $y$-axis, so $y=5$.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

AquaAqua

56.5k1515 gold badges7171 silver badges139139 bronze badges

$endgroup$

add a comment | 

$begingroup$

Consider points $A(2,6)$, $B(-4,3)$ and a point $P(0,y)$ on y-axis. Then notice that given equation is actually $$AP+BP= AB$$ so $P$ is (by triangle inequality ) on a line $AB$ so it is an intersection point of the line $AB$: $$ y= 1over 2x+5$$ and $y$-axis, so $y=5$.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

AquaAqua

56.5k1515 gold badges7171 silver badges139139 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

AquaAqua

56.5k1515 gold badges7171 silver badges139139 bronze badges

answered 7 hours ago

AquaAqua

56.5k1515 gold badges7171 silver badges139139 bronze badges

answered 7 hours ago

AquaAqua

56.5k1515 gold badges7171 silver badges139139 bronze badges

0

$begingroup$

Hint: After squaring one times we get
$$2sqrt4+(y-6)^2sqrt16+(y-3)^2=25-(y-6)^2-(y-3)^2$$
squaring again and simplfying we get
$$4 left(46 y^2-450 y+975right)=0$$

share|cite|improve this answer

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

86.6k44 gold badges2929 silver badges7171 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

Hint: After squaring one times we get
$$2sqrt4+(y-6)^2sqrt16+(y-3)^2=25-(y-6)^2-(y-3)^2$$
squaring again and simplfying we get
$$4 left(46 y^2-450 y+975right)=0$$

share|cite|improve this answer

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

86.6k44 gold badges2929 silver badges7171 bronze badges

$endgroup$

add a comment | 

$begingroup$

Hint: After squaring one times we get
$$2sqrt4+(y-6)^2sqrt16+(y-3)^2=25-(y-6)^2-(y-3)^2$$
squaring again and simplfying we get
$$4 left(46 y^2-450 y+975right)=0$$

share|cite|improve this answer

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

86.6k44 gold badges2929 silver badges7171 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

86.6k44 gold badges2929 silver badges7171 bronze badges

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

86.6k44 gold badges2929 silver badges7171 bronze badges

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

86.6k44 gold badges2929 silver badges7171 bronze badges

0

$begingroup$

You can factor out $2$ at the second summand.

$$sqrtunderbrace4+(y-6)^2_=5+2cdot sqrtunderbrace4+left(fracy-32right)^2_=5=sqrt5+2cdotsqrt5$$

Now we see that the following equations has to be true at the same time.

1. $(y-6)^2=1 Rightarrow y_1=7,y_2=5$




2. $left(fracy-32right)^2=1Rightarrow y_1=5,y_2=1$

Thus the solution is $y=...$

share|cite|improve this answer

answered 7 hours ago

callculuscallculus

20k33 gold badges1717 silver badges3232 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

You can factor out $2$ at the second summand.

$$sqrtunderbrace4+(y-6)^2_=5+2cdot sqrtunderbrace4+left(fracy-32right)^2_=5=sqrt5+2cdotsqrt5$$

Now we see that the following equations has to be true at the same time.

1. $(y-6)^2=1 Rightarrow y_1=7,y_2=5$




2. $left(fracy-32right)^2=1Rightarrow y_1=5,y_2=1$

Thus the solution is $y=...$

share|cite|improve this answer

answered 7 hours ago

callculuscallculus

20k33 gold badges1717 silver badges3232 bronze badges

$endgroup$

add a comment | 

$begingroup$

You can factor out $2$ at the second summand.

$$sqrtunderbrace4+(y-6)^2_=5+2cdot sqrtunderbrace4+left(fracy-32right)^2_=5=sqrt5+2cdotsqrt5$$

Now we see that the following equations has to be true at the same time.

1. $(y-6)^2=1 Rightarrow y_1=7,y_2=5$




2. $left(fracy-32right)^2=1Rightarrow y_1=5,y_2=1$

Thus the solution is $y=...$

share|cite|improve this answer

answered 7 hours ago

callculuscallculus

20k33 gold badges1717 silver badges3232 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

callculuscallculus

20k33 gold badges1717 silver badges3232 bronze badges

answered 7 hours ago

callculuscallculus

20k33 gold badges1717 silver badges3232 bronze badges

answered 7 hours ago

callculuscallculus

20k33 gold badges1717 silver badges3232 bronze badges