Harvard University Math Exam || Algebra Problem Tricks || 99% Failed Admission Aptitude Test
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- Опубликовано: 11 сен 2024
- Harvard University Exam Question || Algebra Problem || Aptitude Simplification Test
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We could factor the original equation into:
(x^2/(x-3))^2 -6(x^2/(x-3)^2)=16.
Then let w= the stuff inside the parentheses, which becomes:
w^2-6w=16
After this, it's easy. You did the same, but subbing this way on the front end eliminates a lot of work.
I enjoy your videos. Thanks for sharing. I must admit, however, I do not get the solution all the time and need a nudge!
Thanks so much 😍. Practice builds perfection.
I think a substitution is easier.
Let y=3x/(x-3), thus xy=3(x+y). The original equation becomes x^2+y^2=16
Square both sides, and we have:
(xy)^2=9(x^2+y^2+2xy)=9*2xy+9*16.
We have a quadratic equation of (xy)
(xy)^2-18(xy)-144=0
This can be solved easily.
xy=24, or xy=-6.
xy=3(x+y)
Thus, we have xy=24 and x+y=8, yielding two complex roots x=4+2sqrt(2)*I, or x=4-2sqrt(2)*I
Or xy=-6 and x+y=-2, yielding two real roots x=-1+sqrt(7), or x=-1-sqrt(7).
Obviously we need to check and see if the absolute value of each real root is smaller than 4, which they are.
Thanks for this nice approach!
I was going to do the same thing, but you beat me to it. Nice.
Everything is easy when you know how to do it.
You can multiply both sides of the equation by the denominator, move everything to one side, and simplify.
You get:
x^4 - 6 x^3 + 2 x^2 + 96 x - 144 = 0
Then, factorize the expression.
(x^2 - 8 x + 24) (x^2 + 2 x - 6) = 0
From there, it's easy
x = -1 - sqrt(7)
x = -1 +sqrt(7)
x = 4 + 2 i sqrt(2)
x = 4 + 2 i sqrt(2)
But.....
Knowing how to get to a solution is often more valuable than the solution itself.
This is why, instead of films that show how to solve a problem, I prefer those that show how to find out how to solve a problem.
Absolutely 💯
How did you factor it?
@@souzasilva5471 This is a fundamental issue... How do we know what to extract from parentheses and how to group it?
I prefer methods rather than tricks.
I assumed I would try to factor the polynomial into the product of two polynomials with integer coefficients. I wrote the equations in a sufficiently general form,
(x^2+c1x+d1)*(x^2+c2x+d2)
multiplied, compared the coefficients with the polynomial being factored, and solved the system of equations with integer solutions.
The last equation is d1⋅d2=d, so the number of divisors of the constant term is significant for the ease of solving the problem.
In this case, we are unlucky; there are many divisors of 144, but it’s enough to find just one solution.
However, you can substitute x=z+1, multiply, and find the factorization in terms of the variable z, and then substitute back into the factorization.
If you’re interested in the details, I can describe them tomorrow. Sometimes you can make a substitution to simplify things, but I expect that the solver will show how they came up with that substitution. I invented the substitution z=x+1 to change the constant term of the polynomial, hoping that this will reduce the number of divisors.
and I was right
z^4 - 2 z^3 - 10 z^2 + 86 z - 51 = 0
51=3*17
The equation to solve is
x² + (3x/(x − 3))² = 16
A different approach which does not require a substitution and which does not require having to deal with fractions is to start by multiplying both sides by (x − 3)² to get rid of the fraction. We can then proceed as follows
x²(x − 3)² + 9x² = 16(x − 3)²
x²(x² − 6x + 9) + 9x² = 16(x − 3)²
x²(x² − 6x + 18) = 16(x − 3)²
(x² − 3(x − 3) + 3(x − 3))(x² − 3(x − 3) − 3(x − 3)) = 16(x − 3)²
(x² − 3x + 9)² − 9(x − 3)² = 16(x − 3)²
(x² − 3x + 9)² − 25(x − 3)² = 0
(x² − 3x + 9 + 5(x − 3))(x² − 3x + 9 − 5(x − 3)) = 0
(x² + 2x − 6)(x² − 8x + 24) = 0
x² + 2x − 6 = 0 ⋁ x² − 8x + 24 = 0
(x + 1)² = 7 ⋁ (x − 4)² = −8
x + 1 = √7 ⋁ x + 1 = −√7 ⋁ x − 4 = 2i√2 ⋁ x − 4 = −2i√2
x = −1 + √7 ⋁ x = −1 − √7 ⋁ x = 4 + 2i√2 ⋁ x = 4 − 2i√2
The critical step here is converting the product x²(x² − 6x + 18) into a difference of two squares. It is always possible to convert a product of two quantities into a difference of two squares by taking the average of the two quantities and half the difference between the two quantities. Then, we can get the original quantities back by adding half the difference to their average and by subtracting half the difference from their average. Finally, we can use the difference of two squares identity to turn the product of the sum and the difference of the average and half the difference into a difference of two squares.
If we have two quantities p and q, then their average is ½(p + q) and half their difference is ½(p − q) and then p = ½(p + q) + ½(p − q) and q = ½(p + q) − ½(p − q) so applying the difference of two squares identity (a + b)(a − b) = a² − b² with a = ½(p + q), b = ½(p − q) we have
p·q = (½(p + q) + ½(p − q))·(½(p + q) − ½(p − q)) = (½(p + q))² − (½(p − q))²
Here, we have the product x²(x² − 6x + 18). The average of x² and x² − 6x + 18 is ½(x² + x² − 6x + 18) = ½(2x² − 6x + 18) = x² − 3x + 9 = x² − 3(x − 3) and half the difference between x² and x² − 6x + 18 is ½(x² − (x² − 6x + 18)) = ½(6x − 18) = 3x − 9 = 3(x − 3) so we have
x² = x² − 3(x − 3) + 3(x − 3)
x² − 6x + 18 = x² − 3(x − 3) − 3(x − 3)
which gives
x²(x² − 6x + 18) = (x² − 3(x − 3))² − (3(x − 3))² = (x² − 3x + 9)² − 9(x − 3)²
Bringing over 16(x − 3)² from the right hand side to the left hand side the right hand side becomes zero while at the left hand side we then have
(x² − 3x + 9)² − 25(x − 3)²
which is again a difference of two squares. This can be factored into two quadratics using the difference to two squares identity a² − b² = (a + b)(a − b). Since the right hand side is zero, we can then apply the zero product property A·B = 0 ⟺ A = 0 ⋁ B = 0 to obtain two quadratic equations which are easily solved.
Yet another approach is as follows
x² + (3x/(x − 3))² = 16
(x − 3)² + 6x − 9 + (3x/(x − 3))² = 16
(x − 3)² + 6x + (3x/(x − 3))² − 25 = 0
((x − 3) + 3x/(x − 3))² − 5² = 0
((x − 3)² + 3x)² − (5(x − 3))² = 0
(x² − 3x + 9)² − (5x − 15)² = 0
x² + 2x − 6 = 0 ⋁ x² − 8x + 24 = 0
(x + 1)² = 7 ⋁ (x − 4)² = −8
x + 1 = √7 ⋁ x + 1 = −√7 ⋁ x − 4 = 2i√2 ⋁ x − 4 = −2i√2
x = −1 + √7 ⋁ x = −1 − √7 ⋁ x = 4 + 2i√2 ⋁ x = 4 − 2i√2
Here, the critical step is rewriting x² as (x − 3)² + 6x − 9 and noting that
(x − 3)² + 6x + (3x/(x − 3))²
is a perfect square since 6x = 2·(x − 3)·(3x/(x − 3)) is twice the product of (x − 3) and (3x/(x − 3)). So, with a = (x − 3), b = (3x/(x − 3)) and applying the identity a² + 2·a·b + b² = (a + b)² we have
(x − 3)² + 6x + (3x/(x − 3))² = ((x − 3) + 3x/(x − 3))²
Bringing over the constant 16 from the right hand side to the left hand side this means that we can write the left hand side of the equation as a difference of two squares
((x − 3) + 3x/(x − 3))² − 5²
while the right hand side is now zero. To eliminate the fraction we then multiply both sides of the equation by (x − 3)² which turns the left hand side into
((x − 3)² + 3x)² − (5(x −3))²
which gives
(x² − 3x + 9)² − (5x − 15)²
Now the left hand side is again a difference of two squares which can be factored into two quadratics using the difference to two squares identity a² − b² = (a + b)(a − b). Since the right hand side is zero, we can then apply the zero product property A·B = 0 ⟺ A = 0 ⋁ B = 0 to obtain two quadratic equations which are easily solved.
Thanks for detailed resourceful explanation 🙏🙏🙏
{x2+x^2 ➖ }=x^4 (9x^2/9x^2)={x^4+9x^2}/9x^2=9x^6/9x^2= 1x^3 (x ➖ 3x+1).
-4^2 +[(3•-4)^2]/-4-3=16+2,93=18,93
[-3,6 >x> -3,7]
-3,65^2 + (3•-3,65/-3,65-3)^2=
13,32 + 2,71= 16,03 x= -3,65
Harvard University Math Exam: x² + [3x/(x - 3)]² = 16; x =?
x ≠ 3; x² + [3x/(x - 3)]² = x²[1 + 9/(x - 3)²] = 16
Let: y = x - 3, x = y + 3; x²[1 + 9/(x - 3)]² = [(y + 3)²](1 + 9/y²) = 16, y ≠ 0
[(y + 3)²](y² + 9) = 16y², (y² + 6y + 9)(y² + 9) = [(y² + 9) + 6y](y² + 9) = 16y²
(y² + 9)² + 6y(y² + 9) - 16y² = 0, (y² + 9 + 8y)(y² + 9 - 2y) = 0
y² + 9 + 8y = y² + 8y + 9 = 0 or y² + 9 - 2y = y² - 2y + 9 = 0
y² + 8y + 9 = (x - 3)² + 8(x - 3) + 9 = x² - 6x + 9 + 8(x - 3) + 9 = x² + 2x - 6 = 0
x² + 2x + 1 = (x + 1)² = 7 = (√7)²; x = - 1 ± √7
y² - 2y + 9 = (x - 3)² - 2(x - 3) + 9 = x² - 6x + 9 - 2x + 6 + 9 = x² - 8x + 24 = 0
x² - 8x + 16 = (x - 4)² = - 8 = (2i√2)²; x = 4 ± 2i√2
Answer check:
x² + [3x/(x - 3)]² = [(y + 3)²](1 + 9/y²); y = x - 3
x = - 1 ± √7 = y + 3, y = - 4 ± √7, y² = (- 4 ± √7)² = 23 -/+ 8√7
[(y + 3)²](1 + 9/y²) = [(- 1 ± √7)²][1 + 9/(23 -/+ 8√7)]
= (8 -/+ 2√7)[(32 -/+ 8√7)/(23 -/+ 8√7)] = [2(4 -/+ √7)][8(4 -/+ √7)/(23 -/+ 8√7)]
= 16[(4 -/+ √7)²]/(23 -/+ 8√7) = 16[(23 -/+ 8√7)/(23 -/+ 8√7)] = 16; Confirmed
x = 4 ± 2i√2 = 2(2 ± i√2) = y + 3, y = 1 ± 2i√2, y² = (1 ± 2i√2)² = - 7 ± 4i√2
[(4(2 ± i√2)²][1 + 9/(- 7 ± 4i√2)] = [8(1 ± 2i√2)][2(1 ± 2i√2)/(- 7 ± 4i√2)]
= [16(1 ± 2i√2)²]/(- 7 ± 4i√2) = 16(1 - 8 ± 4i√2)]/(- 7 ± 4i√2) = 16; Confirmed
Final answer:
x = - 1 + √7; x = - 1 - √7; Two complex value roots, x = 4 + 2i√2 or x = 4 - 2i√2
Many thanks indeed
You're welcome 😊. I'm glad you liked it too
It's easy for japanese highschool student.
I believe that an issue of this level can only be resolved by those who prepared it. Was it you, the teacher, who prepared it?
Anybody can solve it. What you need to know is algebraic identities
Não é tão simples como você coloca. O normal é chegar a uma equação do quarto grau e tentar resolver. (It's not as simple as you put it. The normal thing is to come up with a fourth degree equation and try to solve it.)
Hello,
A the time 15:40, X = (4 ± 2 i √2)÷2, not X=(4±2i√2).
Thank you,
x = 6 is the answer
It is a question of class 9 in Bangladesh
Really 😂
it was in German or English?