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quadratic equation solving mistake
Unfoiling quadratic equationSolve a product of binomials raised to a power for xSolve for “x” and “y”How do pupils solve 2nd degree equations in Germany? (different from Spain)Where did I go wrong in solving this system of equations?Roots of quadratic equation by completing the square or other method?New way to solve Quadratic Equations?The solution set of a multivariate quadratic after transformation of variablesSame equation, different results (Quartic function)What is rule for when solving algebraic equations?
$begingroup$
I'm a student who started self learning quadratic equations for a youth university program. I'm busy at trying to solve such equation:
$$
(1 - 4x)^2 + 9x + 7 = 2(x+3)(1-x) + (x+4)^2
$$
this is my current progress:
beginalign
(1 - 4x)^2 + 9x + 7 &= 2(x+3)(1-x)+ (x+4)^2\
(1 - 4x)(1 - 4x) + 9x + 7 &= (2x + 6)(1 - x) + (x + 4)(x + 4)\
1 - 4x - 4x + 16x^2 + 9x + 7 &= 2x - 2x^2 + 6 - 6x + x^2 + 4x + 4x + 16\
8 + 16x^2 + x &= 2x - x^2 + 6 - 6x + 8x + 16\
8 + 16x^2 + x &= 4x - x^2 + 22\
16x^2 + x &= 4x - x^2 + 14\
16x^2 &= 3x - x^2 + 14\
17x^2 &= 3x + 14
endalign
The solutions to this equation are $x = 1,~x=-14/17$.
So, where is my mistake? $x$ is negative, so I must be incorrect.
algebra-precalculus quadratics
edited yesterday
Jam
5,0742 gold badges14 silver badges32 bronze badges
asked Jun 21 at 10:41
neo5003neo5003
313 bronze badges
$endgroup$
add a comment |
$begingroup$
I'm a student who started self learning quadratic equations for a youth university program. I'm busy at trying to solve such equation:
$$
(1 - 4x)^2 + 9x + 7 = 2(x+3)(1-x) + (x+4)^2
$$
this is my current progress:
beginalign
(1 - 4x)^2 + 9x + 7 &= 2(x+3)(1-x)+ (x+4)^2\
(1 - 4x)(1 - 4x) + 9x + 7 &= (2x + 6)(1 - x) + (x + 4)(x + 4)\
1 - 4x - 4x + 16x^2 + 9x + 7 &= 2x - 2x^2 + 6 - 6x + x^2 + 4x + 4x + 16\
8 + 16x^2 + x &= 2x - x^2 + 6 - 6x + 8x + 16\
8 + 16x^2 + x &= 4x - x^2 + 22\
16x^2 + x &= 4x - x^2 + 14\
16x^2 &= 3x - x^2 + 14\
17x^2 &= 3x + 14
endalign
The solutions to this equation are $x = 1,~x=-14/17$.
So, where is my mistake? $x$ is negative, so I must be incorrect.
algebra-precalculus quadratics
edited yesterday
Jam
5,0742 gold badges14 silver badges32 bronze badges
asked Jun 21 at 10:41
neo5003neo5003
313 bronze badges
$endgroup$
2
$begingroup$
Why do you think the result must be wrong ? It is correct ! You should however write $x_1=1$ , $x_2=-frac1417$ to avoid confusion.
$endgroup$
– Peter
Jun 21 at 10:53
4
$begingroup$
When you say the solution is $x = 1/(-14/17)$, what do you mean, exactly? If you meant "the solutions are $x = 1$ or $x = -14/17$", then you should say it that way, as $/$ most commonly means division, and shouldn't be used to signify "or" in math expressions.
$endgroup$
– Arthur
Jun 21 at 10:53
add a comment |
$begingroup$
I'm a student who started self learning quadratic equations for a youth university program. I'm busy at trying to solve such equation:
$$
(1 - 4x)^2 + 9x + 7 = 2(x+3)(1-x) + (x+4)^2
$$
this is my current progress:
beginalign
(1 - 4x)^2 + 9x + 7 &= 2(x+3)(1-x)+ (x+4)^2\
(1 - 4x)(1 - 4x) + 9x + 7 &= (2x + 6)(1 - x) + (x + 4)(x + 4)\
1 - 4x - 4x + 16x^2 + 9x + 7 &= 2x - 2x^2 + 6 - 6x + x^2 + 4x + 4x + 16\
8 + 16x^2 + x &= 2x - x^2 + 6 - 6x + 8x + 16\
8 + 16x^2 + x &= 4x - x^2 + 22\
16x^2 + x &= 4x - x^2 + 14\
16x^2 &= 3x - x^2 + 14\
17x^2 &= 3x + 14
endalign
The solutions to this equation are $x = 1,~x=-14/17$.
So, where is my mistake? $x$ is negative, so I must be incorrect.
algebra-precalculus quadratics
edited yesterday
Jam
5,0742 gold badges14 silver badges32 bronze badges
asked Jun 21 at 10:41
neo5003neo5003
313 bronze badges
$endgroup$
I'm a student who started self learning quadratic equations for a youth university program. I'm busy at trying to solve such equation:
$$
(1 - 4x)^2 + 9x + 7 = 2(x+3)(1-x) + (x+4)^2
$$
this is my current progress:
beginalign
(1 - 4x)^2 + 9x + 7 &= 2(x+3)(1-x)+ (x+4)^2\
(1 - 4x)(1 - 4x) + 9x + 7 &= (2x + 6)(1 - x) + (x + 4)(x + 4)\
1 - 4x - 4x + 16x^2 + 9x + 7 &= 2x - 2x^2 + 6 - 6x + x^2 + 4x + 4x + 16\
8 + 16x^2 + x &= 2x - x^2 + 6 - 6x + 8x + 16\
8 + 16x^2 + x &= 4x - x^2 + 22\
16x^2 + x &= 4x - x^2 + 14\
16x^2 &= 3x - x^2 + 14\
17x^2 &= 3x + 14
endalign
The solutions to this equation are $x = 1,~x=-14/17$.
So, where is my mistake? $x$ is negative, so I must be incorrect.
algebra-precalculus quadratics
algebra-precalculus quadratics
edited yesterday
Jam
5,0742 gold badges14 silver badges32 bronze badges
asked Jun 21 at 10:41
neo5003neo5003
313 bronze badges
edited yesterday
Jam
5,0742 gold badges14 silver badges32 bronze badges
asked Jun 21 at 10:41
neo5003neo5003
313 bronze badges
edited yesterday
Jam
5,0742 gold badges14 silver badges32 bronze badges
edited yesterday
Jam
5,0742 gold badges14 silver badges32 bronze badges
edited yesterday
Jam
5,0742 gold badges14 silver badges32 bronze badges
5,0742 gold badges14 silver badges32 bronze badges
asked Jun 21 at 10:41
neo5003neo5003
313 bronze badges
asked Jun 21 at 10:41
neo5003neo5003
313 bronze badges
asked Jun 21 at 10:41
neo5003neo5003
313 bronze badges
313 bronze badges
2
$begingroup$
Why do you think the result must be wrong ? It is correct ! You should however write $x_1=1$ , $x_2=-frac1417$ to avoid confusion.
$endgroup$
– Peter
Jun 21 at 10:53
4
$begingroup$
When you say the solution is $x = 1/(-14/17)$, what do you mean, exactly? If you meant "the solutions are $x = 1$ or $x = -14/17$", then you should say it that way, as $/$ most commonly means division, and shouldn't be used to signify "or" in math expressions.
$endgroup$
– Arthur
Jun 21 at 10:53
add a comment |
2
$begingroup$
Why do you think the result must be wrong ? It is correct ! You should however write $x_1=1$ , $x_2=-frac1417$ to avoid confusion.
$endgroup$
– Peter
Jun 21 at 10:53
4
$begingroup$
When you say the solution is $x = 1/(-14/17)$, what do you mean, exactly? If you meant "the solutions are $x = 1$ or $x = -14/17$", then you should say it that way, as $/$ most commonly means division, and shouldn't be used to signify "or" in math expressions.
$endgroup$
– Arthur
Jun 21 at 10:53
2
2
$begingroup$
Why do you think the result must be wrong ? It is correct ! You should however write $x_1=1$ , $x_2=-frac1417$ to avoid confusion.
$endgroup$
– Peter
Jun 21 at 10:53
$begingroup$
Why do you think the result must be wrong ? It is correct ! You should however write $x_1=1$ , $x_2=-frac1417$ to avoid confusion.
$endgroup$
– Peter
Jun 21 at 10:53
4
4
$begingroup$
When you say the solution is $x = 1/(-14/17)$, what do you mean, exactly? If you meant "the solutions are $x = 1$ or $x = -14/17$", then you should say it that way, as $/$ most commonly means division, and shouldn't be used to signify "or" in math expressions.
$endgroup$
– Arthur
Jun 21 at 10:53
$begingroup$
When you say the solution is $x = 1/(-14/17)$, what do you mean, exactly? If you meant "the solutions are $x = 1$ or $x = -14/17$", then you should say it that way, as $/$ most commonly means division, and shouldn't be used to signify "or" in math expressions.
$endgroup$
– Arthur
Jun 21 at 10:53
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your working is all fine. Only the last step was remaining to be written down after factoring.
answered Jun 21 at 11:32
NarasimhamNarasimham
21.7k6 gold badges22 silver badges58 bronze badges
$endgroup$
add a comment |
$begingroup$
You have not made any mistake.
The final equation you have obtained is
$$17x^2-3x-14=0$$
$$17x^2-17x+14x-14=0$$
$$17x(x-1)+14(x-1)=0$$
$$(17x+14)(x-1)=0$$
which has the roots $1$ and $frac-1417$.
answered Jun 21 at 10:54
EpiksaladEpiksalad
9632 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
$$
(1 - 4)^2 + 9 + 7 = 2(1+3)(1-1) + (1+4)^2
$$
seems true ($25$ in both members), and with a little more effort
$$
left(1 - 4fracoverline1417right)^2 + 9fracoverline1417 + 7 = 2left(fracoverline1417+3right)left(1-fracoverline1417right) + left(fracoverline1417+4right)^2
$$
is
$$(17+56)^2-9cdot17cdot14+7cdot17^2=2(-14+51)(17+14)+(-14+68)^2$$which is $5210$ on both sides.
As a quadratic equation has at most two roots, your work is right.
answered yesterday
Yves DaoustYves Daoust
139k8 gold badges82 silver badges239 bronze badges
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your working is all fine. Only the last step was remaining to be written down after factoring.
answered Jun 21 at 11:32
NarasimhamNarasimham
21.7k6 gold badges22 silver badges58 bronze badges
$endgroup$
add a comment |
$begingroup$
Your working is all fine. Only the last step was remaining to be written down after factoring.
answered Jun 21 at 11:32
NarasimhamNarasimham
21.7k6 gold badges22 silver badges58 bronze badges
$endgroup$
add a comment |
$begingroup$
Your working is all fine. Only the last step was remaining to be written down after factoring.
answered Jun 21 at 11:32
NarasimhamNarasimham
21.7k6 gold badges22 silver badges58 bronze badges
$endgroup$
Your working is all fine. Only the last step was remaining to be written down after factoring.
answered Jun 21 at 11:32
NarasimhamNarasimham
21.7k6 gold badges22 silver badges58 bronze badges
answered Jun 21 at 11:32
NarasimhamNarasimham
21.7k6 gold badges22 silver badges58 bronze badges
answered Jun 21 at 11:32
NarasimhamNarasimham
21.7k6 gold badges22 silver badges58 bronze badges
answered Jun 21 at 11:32
NarasimhamNarasimham
21.7k6 gold badges22 silver badges58 bronze badges
21.7k6 gold badges22 silver badges58 bronze badges
add a comment |
add a comment |
$begingroup$
You have not made any mistake.
The final equation you have obtained is
$$17x^2-3x-14=0$$
$$17x^2-17x+14x-14=0$$
$$17x(x-1)+14(x-1)=0$$
$$(17x+14)(x-1)=0$$
which has the roots $1$ and $frac-1417$.
answered Jun 21 at 10:54
EpiksaladEpiksalad
9632 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
You have not made any mistake.
The final equation you have obtained is
$$17x^2-3x-14=0$$
$$17x^2-17x+14x-14=0$$
$$17x(x-1)+14(x-1)=0$$
$$(17x+14)(x-1)=0$$
which has the roots $1$ and $frac-1417$.
answered Jun 21 at 10:54
EpiksaladEpiksalad
9632 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
You have not made any mistake.
The final equation you have obtained is
$$17x^2-3x-14=0$$
$$17x^2-17x+14x-14=0$$
$$17x(x-1)+14(x-1)=0$$
$$(17x+14)(x-1)=0$$
which has the roots $1$ and $frac-1417$.
answered Jun 21 at 10:54
EpiksaladEpiksalad
9632 silver badges12 bronze badges
$endgroup$
You have not made any mistake.
The final equation you have obtained is
$$17x^2-3x-14=0$$
$$17x^2-17x+14x-14=0$$
$$17x(x-1)+14(x-1)=0$$
$$(17x+14)(x-1)=0$$
which has the roots $1$ and $frac-1417$.
answered Jun 21 at 10:54
EpiksaladEpiksalad
9632 silver badges12 bronze badges
answered Jun 21 at 10:54
EpiksaladEpiksalad
9632 silver badges12 bronze badges
answered Jun 21 at 10:54
EpiksaladEpiksalad
9632 silver badges12 bronze badges
answered Jun 21 at 10:54
EpiksaladEpiksalad
9632 silver badges12 bronze badges
9632 silver badges12 bronze badges
add a comment |
add a comment |
$begingroup$
$$
(1 - 4)^2 + 9 + 7 = 2(1+3)(1-1) + (1+4)^2
$$
seems true ($25$ in both members), and with a little more effort
$$
left(1 - 4fracoverline1417right)^2 + 9fracoverline1417 + 7 = 2left(fracoverline1417+3right)left(1-fracoverline1417right) + left(fracoverline1417+4right)^2
$$
is
$$(17+56)^2-9cdot17cdot14+7cdot17^2=2(-14+51)(17+14)+(-14+68)^2$$which is $5210$ on both sides.
As a quadratic equation has at most two roots, your work is right.
answered yesterday
Yves DaoustYves Daoust
139k8 gold badges82 silver badges239 bronze badges
$endgroup$
add a comment |
$begingroup$
$$
(1 - 4)^2 + 9 + 7 = 2(1+3)(1-1) + (1+4)^2
$$
seems true ($25$ in both members), and with a little more effort
$$
left(1 - 4fracoverline1417right)^2 + 9fracoverline1417 + 7 = 2left(fracoverline1417+3right)left(1-fracoverline1417right) + left(fracoverline1417+4right)^2
$$
is
$$(17+56)^2-9cdot17cdot14+7cdot17^2=2(-14+51)(17+14)+(-14+68)^2$$which is $5210$ on both sides.
As a quadratic equation has at most two roots, your work is right.
answered yesterday
Yves DaoustYves Daoust
139k8 gold badges82 silver badges239 bronze badges
$endgroup$
add a comment |
$begingroup$
$$
(1 - 4)^2 + 9 + 7 = 2(1+3)(1-1) + (1+4)^2
$$
seems true ($25$ in both members), and with a little more effort
$$
left(1 - 4fracoverline1417right)^2 + 9fracoverline1417 + 7 = 2left(fracoverline1417+3right)left(1-fracoverline1417right) + left(fracoverline1417+4right)^2
$$
is
$$(17+56)^2-9cdot17cdot14+7cdot17^2=2(-14+51)(17+14)+(-14+68)^2$$which is $5210$ on both sides.
As a quadratic equation has at most two roots, your work is right.
answered yesterday
Yves DaoustYves Daoust
139k8 gold badges82 silver badges239 bronze badges
$endgroup$
$$
(1 - 4)^2 + 9 + 7 = 2(1+3)(1-1) + (1+4)^2
$$
seems true ($25$ in both members), and with a little more effort
$$
left(1 - 4fracoverline1417right)^2 + 9fracoverline1417 + 7 = 2left(fracoverline1417+3right)left(1-fracoverline1417right) + left(fracoverline1417+4right)^2
$$
is
$$(17+56)^2-9cdot17cdot14+7cdot17^2=2(-14+51)(17+14)+(-14+68)^2$$which is $5210$ on both sides.
As a quadratic equation has at most two roots, your work is right.
answered yesterday
Yves DaoustYves Daoust
139k8 gold badges82 silver badges239 bronze badges
answered yesterday
Yves DaoustYves Daoust
139k8 gold badges82 silver badges239 bronze badges
answered yesterday
Yves DaoustYves Daoust
139k8 gold badges82 silver badges239 bronze badges
answered yesterday
Yves DaoustYves Daoust
139k8 gold badges82 silver badges239 bronze badges
139k8 gold badges82 silver badges239 bronze badges
add a comment |
add a comment |
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$begingroup$
Why do you think the result must be wrong ? It is correct ! You should however write $x_1=1$ , $x_2=-frac1417$ to avoid confusion.
$endgroup$
– Peter
Jun 21 at 10:53
4
$begingroup$
When you say the solution is $x = 1/(-14/17)$, what do you mean, exactly? If you meant "the solutions are $x = 1$ or $x = -14/17$", then you should say it that way, as $/$ most commonly means division, and shouldn't be used to signify "or" in math expressions.
$endgroup$
– Arthur
Jun 21 at 10:53