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Solve $frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=0.8$
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up vote
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Solve:
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here except that there's some similiraty between some pairs of denominators, my intuation is telling me to just multipy everything and try to solve but it looks like a big dead polynomial.
Is there an elegent way to solve this?
Solution:
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=0.8$$
$$frac1x+1-frac1x+5=0.8$$
$$4=0.8(x+1)(x+5)$$
$$4x^2+24x=0$$
$$4x(x+6)=0$$
Solution: $x_1=0$ and $x_2=-6$
Definately very elegent! :)
polynomials
asked Jul 17 at 13:34
Maxim
616314
add a comment |Â
up vote
6
down vote
favorite
Solve:
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here except that there's some similiraty between some pairs of denominators, my intuation is telling me to just multipy everything and try to solve but it looks like a big dead polynomial.
Is there an elegent way to solve this?
Solution:
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=0.8$$
$$frac1x+1-frac1x+5=0.8$$
$$4=0.8(x+1)(x+5)$$
$$4x^2+24x=0$$
$$4x(x+6)=0$$
Solution: $x_1=0$ and $x_2=-6$
Definately very elegent! :)
polynomials
asked Jul 17 at 13:34
Maxim
616314
9
Hint: $frac1(x+k)(x+k+1) = frac1x+k - frac1x+k+1$
– achille hui
Jul 17 at 13:37
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Solve:
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here except that there's some similiraty between some pairs of denominators, my intuation is telling me to just multipy everything and try to solve but it looks like a big dead polynomial.
Is there an elegent way to solve this?
Solution:
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=0.8$$
$$frac1x+1-frac1x+5=0.8$$
$$4=0.8(x+1)(x+5)$$
$$4x^2+24x=0$$
$$4x(x+6)=0$$
Solution: $x_1=0$ and $x_2=-6$
Definately very elegent! :)
polynomials
asked Jul 17 at 13:34
Maxim
616314
Solve:
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here except that there's some similiraty between some pairs of denominators, my intuation is telling me to just multipy everything and try to solve but it looks like a big dead polynomial.
Is there an elegent way to solve this?
Solution:
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=0.8$$
$$frac1x+1-frac1x+5=0.8$$
$$4=0.8(x+1)(x+5)$$
$$4x^2+24x=0$$
$$4x(x+6)=0$$
Solution: $x_1=0$ and $x_2=-6$
Definately very elegent! :)
polynomials
asked Jul 17 at 13:34
Maxim
616314
edited Jul 17 at 13:53
asked Jul 17 at 13:34
Maxim
616314
asked Jul 17 at 13:34
Maxim
616314
asked Jul 17 at 13:34
Maxim
616314
616314
9
Hint: $frac1(x+k)(x+k+1) = frac1x+k - frac1x+k+1$
– achille hui
Jul 17 at 13:37
add a comment |Â
9
Hint: $frac1(x+k)(x+k+1) = frac1x+k - frac1x+k+1$
– achille hui
Jul 17 at 13:37
9
9
Hint: $frac1(x+k)(x+k+1) = frac1x+k - frac1x+k+1$
– achille hui
Jul 17 at 13:37
Hint: $frac1(x+k)(x+k+1) = frac1x+k - frac1x+k+1$
– achille hui
Jul 17 at 13:37
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
7
down vote
accepted
HINT
We have that
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=$$
$$=frac1(x+1)colorred-frac1(x+2)+frac1(x+2)-frac1(x+3)+frac1(x+3)-frac1(x+4)+frac1(x+4)-frac1(x+5)$$
answered Jul 17 at 13:39
gimusi
65.4k73584
add a comment |Â
up vote
6
down vote
Make the equations more symmetric by shifting $z:=x+3$ and simplify
$$frac1(z-2)(z-1)+frac1(z-1)z+frac1z(z+1)+frac1(z+1)(z+2)$$
$$=frac2z^2+4(z^2-4)(z^2-1)+frac2z^2-1$$
$$=frac4z^2-4(z^2-4)(z^2-1)$$
$$=frac4z^2-4$$
giving $z=pm 3$ and $x=0,x=-6$.
answered Jul 17 at 13:47
Yves Daoust
111k665204
Strange... I got a bit of a different result.. Am I wrong in my solutions? (I've edited my question)
– Maxim
Jul 17 at 13:57
1
@Maxim: there was a typo in the last result.
– Yves Daoust
Jul 17 at 14:00
add a comment |Â
up vote
3
down vote
Hint: your equation is equivalent to $$frac4x(6+x)5(1+x)(5+x)=0$$
answered Jul 17 at 13:54
Dr. Sonnhard Graubner
66.9k32659
add a comment |Â
up vote
2
down vote
With factors following an arithmetic progression, you establish the rule
$$frac1a(a+b)+frac1(a+b)(a+2b)=frac2a+2ba(a+b)(a+2b)=frac2a(a+2b).$$
Applying it three times, the sum reduces to
$$frac4(x+1)(x+5),$$ giving the easy quadratic equation
$$(x+1)(x+5)=frac40.8$$
or $$x(x+6)=0.$$
answered Jul 17 at 14:00
Yves Daoust
111k665204
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
HINT
We have that
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=$$
$$=frac1(x+1)colorred-frac1(x+2)+frac1(x+2)-frac1(x+3)+frac1(x+3)-frac1(x+4)+frac1(x+4)-frac1(x+5)$$
answered Jul 17 at 13:39
gimusi
65.4k73584
add a comment |Â
up vote
7
down vote
accepted
HINT
We have that
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=$$
$$=frac1(x+1)colorred-frac1(x+2)+frac1(x+2)-frac1(x+3)+frac1(x+3)-frac1(x+4)+frac1(x+4)-frac1(x+5)$$
answered Jul 17 at 13:39
gimusi
65.4k73584
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
HINT
We have that
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=$$
$$=frac1(x+1)colorred-frac1(x+2)+frac1(x+2)-frac1(x+3)+frac1(x+3)-frac1(x+4)+frac1(x+4)-frac1(x+5)$$
answered Jul 17 at 13:39
gimusi
65.4k73584
HINT
We have that
$$frac1(x+1)(x+2)+frac1(x+2)(x+3)+frac1(x+3)(x+4)+frac1(x+4)(x+5)=$$
$$=frac1(x+1)colorred-frac1(x+2)+frac1(x+2)-frac1(x+3)+frac1(x+3)-frac1(x+4)+frac1(x+4)-frac1(x+5)$$
answered Jul 17 at 13:39
gimusi
65.4k73584
answered Jul 17 at 13:39
gimusi
65.4k73584
answered Jul 17 at 13:39
gimusi
65.4k73584
answered Jul 17 at 13:39
gimusi
65.4k73584
65.4k73584
add a comment |Â
add a comment |Â
up vote
6
down vote
Make the equations more symmetric by shifting $z:=x+3$ and simplify
$$frac1(z-2)(z-1)+frac1(z-1)z+frac1z(z+1)+frac1(z+1)(z+2)$$
$$=frac2z^2+4(z^2-4)(z^2-1)+frac2z^2-1$$
$$=frac4z^2-4(z^2-4)(z^2-1)$$
$$=frac4z^2-4$$
giving $z=pm 3$ and $x=0,x=-6$.
answered Jul 17 at 13:47
Yves Daoust
111k665204
Strange... I got a bit of a different result.. Am I wrong in my solutions? (I've edited my question)
– Maxim
Jul 17 at 13:57
1
@Maxim: there was a typo in the last result.
– Yves Daoust
Jul 17 at 14:00
add a comment |Â
up vote
6
down vote
Make the equations more symmetric by shifting $z:=x+3$ and simplify
$$frac1(z-2)(z-1)+frac1(z-1)z+frac1z(z+1)+frac1(z+1)(z+2)$$
$$=frac2z^2+4(z^2-4)(z^2-1)+frac2z^2-1$$
$$=frac4z^2-4(z^2-4)(z^2-1)$$
$$=frac4z^2-4$$
giving $z=pm 3$ and $x=0,x=-6$.
answered Jul 17 at 13:47
Yves Daoust
111k665204
Strange... I got a bit of a different result.. Am I wrong in my solutions? (I've edited my question)
– Maxim
Jul 17 at 13:57
1
@Maxim: there was a typo in the last result.
– Yves Daoust
Jul 17 at 14:00
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Make the equations more symmetric by shifting $z:=x+3$ and simplify
$$frac1(z-2)(z-1)+frac1(z-1)z+frac1z(z+1)+frac1(z+1)(z+2)$$
$$=frac2z^2+4(z^2-4)(z^2-1)+frac2z^2-1$$
$$=frac4z^2-4(z^2-4)(z^2-1)$$
$$=frac4z^2-4$$
giving $z=pm 3$ and $x=0,x=-6$.
answered Jul 17 at 13:47
Yves Daoust
111k665204
Make the equations more symmetric by shifting $z:=x+3$ and simplify
$$frac1(z-2)(z-1)+frac1(z-1)z+frac1z(z+1)+frac1(z+1)(z+2)$$
$$=frac2z^2+4(z^2-4)(z^2-1)+frac2z^2-1$$
$$=frac4z^2-4(z^2-4)(z^2-1)$$
$$=frac4z^2-4$$
giving $z=pm 3$ and $x=0,x=-6$.
answered Jul 17 at 13:47
Yves Daoust
111k665204
edited Jul 17 at 13:59
answered Jul 17 at 13:47
Yves Daoust
111k665204
answered Jul 17 at 13:47
Yves Daoust
111k665204
answered Jul 17 at 13:47
Yves Daoust
111k665204
111k665204
Strange... I got a bit of a different result.. Am I wrong in my solutions? (I've edited my question)
– Maxim
Jul 17 at 13:57
1
@Maxim: there was a typo in the last result.
– Yves Daoust
Jul 17 at 14:00
add a comment |Â
Strange... I got a bit of a different result.. Am I wrong in my solutions? (I've edited my question)
– Maxim
Jul 17 at 13:57
1
@Maxim: there was a typo in the last result.
– Yves Daoust
Jul 17 at 14:00
Strange... I got a bit of a different result.. Am I wrong in my solutions? (I've edited my question)
– Maxim
Jul 17 at 13:57
Strange... I got a bit of a different result.. Am I wrong in my solutions? (I've edited my question)
– Maxim
Jul 17 at 13:57
1
1
@Maxim: there was a typo in the last result.
– Yves Daoust
Jul 17 at 14:00
@Maxim: there was a typo in the last result.
– Yves Daoust
Jul 17 at 14:00
add a comment |Â
up vote
3
down vote
Hint: your equation is equivalent to $$frac4x(6+x)5(1+x)(5+x)=0$$
answered Jul 17 at 13:54
Dr. Sonnhard Graubner
66.9k32659
add a comment |Â
up vote
3
down vote
Hint: your equation is equivalent to $$frac4x(6+x)5(1+x)(5+x)=0$$
answered Jul 17 at 13:54
Dr. Sonnhard Graubner
66.9k32659
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Hint: your equation is equivalent to $$frac4x(6+x)5(1+x)(5+x)=0$$
answered Jul 17 at 13:54
Dr. Sonnhard Graubner
66.9k32659
Hint: your equation is equivalent to $$frac4x(6+x)5(1+x)(5+x)=0$$
answered Jul 17 at 13:54
Dr. Sonnhard Graubner
66.9k32659
answered Jul 17 at 13:54
Dr. Sonnhard Graubner
66.9k32659
answered Jul 17 at 13:54
Dr. Sonnhard Graubner
66.9k32659
answered Jul 17 at 13:54
Dr. Sonnhard Graubner
66.9k32659
66.9k32659
add a comment |Â
add a comment |Â
up vote
2
down vote
With factors following an arithmetic progression, you establish the rule
$$frac1a(a+b)+frac1(a+b)(a+2b)=frac2a+2ba(a+b)(a+2b)=frac2a(a+2b).$$
Applying it three times, the sum reduces to
$$frac4(x+1)(x+5),$$ giving the easy quadratic equation
$$(x+1)(x+5)=frac40.8$$
or $$x(x+6)=0.$$
answered Jul 17 at 14:00
Yves Daoust
111k665204
add a comment |Â
up vote
2
down vote
With factors following an arithmetic progression, you establish the rule
$$frac1a(a+b)+frac1(a+b)(a+2b)=frac2a+2ba(a+b)(a+2b)=frac2a(a+2b).$$
Applying it three times, the sum reduces to
$$frac4(x+1)(x+5),$$ giving the easy quadratic equation
$$(x+1)(x+5)=frac40.8$$
or $$x(x+6)=0.$$
answered Jul 17 at 14:00
Yves Daoust
111k665204
add a comment |Â
up vote
2
down vote
up vote
2
down vote
With factors following an arithmetic progression, you establish the rule
$$frac1a(a+b)+frac1(a+b)(a+2b)=frac2a+2ba(a+b)(a+2b)=frac2a(a+2b).$$
Applying it three times, the sum reduces to
$$frac4(x+1)(x+5),$$ giving the easy quadratic equation
$$(x+1)(x+5)=frac40.8$$
or $$x(x+6)=0.$$
answered Jul 17 at 14:00
Yves Daoust
111k665204
With factors following an arithmetic progression, you establish the rule
$$frac1a(a+b)+frac1(a+b)(a+2b)=frac2a+2ba(a+b)(a+2b)=frac2a(a+2b).$$
Applying it three times, the sum reduces to
$$frac4(x+1)(x+5),$$ giving the easy quadratic equation
$$(x+1)(x+5)=frac40.8$$
or $$x(x+6)=0.$$
answered Jul 17 at 14:00
Yves Daoust
111k665204
edited Jul 17 at 15:43
answered Jul 17 at 14:00
Yves Daoust
111k665204
answered Jul 17 at 14:00
Yves Daoust
111k665204
answered Jul 17 at 14:00
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
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9
Hint: $frac1(x+k)(x+k+1) = frac1x+k - frac1x+k+1$
– achille hui
Jul 17 at 13:37