Algebra discussion

"Tagent" means that the graphs intersect at only one point (the slope at the tangent point of the two graphs is the same).

Yes, that's the purpose of the question.

I present 2 ways to do it (the answer is c=-1).

Method 1: Solve the equation

because there is only one point of tangent of the graph this means that the y-values ​​of both graphs at tagent point are the same, so x² = 2x+c we find c such that x² = 2x+c has only one solution.

x²-2x-c = 0 -> x²-2x = c -> x²-2x+1 = c+1 -> (x-1)² = c+1

So the equation has only one solution when c+1=0 -> c=-1.

Method 2: Calculus

The slope of the parabola at the tangent is 2, which is equal to the slope of the line. Since dy/dx(y=x²) = 2x, at tangent 2x=2 -> x=1, tangent (x, x²) = (1, 1)

We get that the line y = 2x+c passes through the point (1,1), so 1=2(1)+c -> c=-1.

This was challenging for me.

Solution.

Draw graphs by photomath.

Problem 6:

Let a, b be the roots of the equation x²+x-5=0. Create a quadratic equation with the following roots.

1) 2a and 2b

2) a² and b²

Hint : by Vieta's Formula, the equation ax²+bx+c=0 has root sum of -b/a, root product of c/a.

Let a,b,c,d be positive real numbers such that a+b+c+d=1

What is the minimum value of:

(1/a)+(4/b)+(9/c)+(16/d)

I saw all the problems will try to find solution tomorrow.

Note: only partial proof

The HM-GM-AM inequality states that "harmonic mean ≤ geometric mean ≤ arithmetic mean".

For a₁,a₂,...,aₙ to be any positive real number, we get.

n/[(1/a₁)+(1/a₂)+...+(1/aₙ)] ≤ root_n(a₁a₂...aₙ) ≤ (a₁+a₂+...+aₙ)/n

The inequality will be an equation when a₁=a₂=...=aₙ

By looking at the AM-GM inequality a₁=a, a₂=b, a₃=c and a₄=d, we get

(a+b+c+d)/4 ≥ root_4(abcd)

1/4 ≥ root_4(abcd)

abcd ≤ 1/256

1/abcd ≥ 256

576/abcd ≥ 256×576 = 147,456 ...[#]

By looking at the AM-GM inequality a₁=1/a, a₂=4/b, a₃=9/c and a₄=16/d, we get

[(1/a)+(4/b)+(9/c)+(16/d)]/4 ≥ root_4[(1/a)(4/b)(9/c)(16/d) ]

[(1/a)+(4/b)+(9/c)+(16/d)]/4 ≥ root_4[(576/abcd)]

From equation [#], we get [(1/a)+(4/b)+(9/c)+(16/d)]/4 ≥ root_4(147,456).

(1/a)+(4/b)+(9/c)+(16/d) ≥ 4 root_4(147,456) = 32 root(6)

So (1/a)+(4/b)+(9/c)+(16/d) ≥ 32 root(6) = 78.38

However, this still doesn't prove that 78.38 is the minimum. I still can't figure out how to do it. I substituted a=0.1,b=0.2,c=0.3,d=0.4 and found that (1/a)+(4/b)+(9/c)+(16/d) = 100 and I haven't found a lower value yet.

So I only partially solved this problem, i.e. 78.38 ≤ min[(1/a)+(4/b)+(9/c)+(16/d)] ≤ 100.

My hint for you finishing the solution is that the answer is 100

I still don't have a way to prove that 100 is the minimum. I need solution now.

What I can do with proof is that there's no way the minimum is less than 78.38, but that's still not 100.

This is about number theory.

Since x-y is divisible by 2, then there is a ∈ Z such that x-y = 2a, so x+y = 2a+2y = 2(a+y).

Show that x²-y² = (x-y)(x+y) = 2a(2(a+y)) = 4a(a+y), so x²-y² is divisible by 4.

This is about number theory.

Definition: Let S={a₁,a₂,a₃,...,aₙ} We call S a "reduced residual system in modulo n" if the remainder dividing a₁,a₂,a₃,...,aₙ by n all different.

Consider {n, n+1} is a reduced residue system in modulo 2, thus 2 | n(n+1).

{n, n+1, n+5} is a reduced residue system in modulo 3, thus 3 | n(n+1).

So it can be concluded that 6 | n(n+1)(n+5) for all n ∈ Z.

15) The equation ax²+bx+c has repeated roots if b²-4ac = 0, so (2(1+3m))² - 4(1)7(3+2m) = 0

36m²+24m+4-84-56m = 0 = 36m²-32m-80 = 4(9m+10)(m-2)

So m = -10/9 or 2

16) From the problem, (x²-bx)(m+1) = (ax-c)(m-1), that is [(m+1)x²+(-bm-b)x] - [(am-a)x+(c-cm)] = 0

(m+1)x² + (-bm-b-am+a)x + (cm-c) = 0

Since the root sum equals 0, for the equation ax²+bx+c = 0, -b/a = 0, b=0.

So -bm-b-am+a = 0 that is (-a-b)m-(b-a) = 0.

-(a+b)m = b-a -> m=(a-b)/(a+b) and m+1≠0 because the right side of the equation is invalid.

So m=(a-b)/(a+b) where a≠0

The equation ax²+bx+c has rational roots if b²-4ac is the absolute square.

17.1) (a+c-b)² + 2cx + (b+c-a) = 0 has a discreminant

(2c)² - 4(a+c-b)(b+c-a) = 4c² - 4(c+(a-b))(c-(a-b)) = 4c² - 4(c²-(a-b)²) = 4c²-4c²+. 4(a-b)² = (2(a-b))² is an absolute square.

17.2) has not been solved.

By Vieta's Formula, the equation ax²+bx+c=0 has root sum of -b/a, root product of c/a.

Let p,q be the roots of the equation. 2x²+2(m+n)x+(m²+n²) = 0 so p+q=-m-n and pq=(m²+n²)/2.

Consider an equation whose roots are (p+q)², (p-q)².

There is a root sum of 2p²+2q² = 2(p+q)²-4pq = 2(-m-n)²-2(m²+n²) = 4mn.

There is a root product of (p²-q²)² = p⁴-2p²q²+q⁴ = (p²+q²)²-4p²q² = (2mn)² - 4[(m²+n²)/2]² = 4m²n² - (m²+n²)² = -(m²-n²)²

Thus, an equation with roots of (p+q)², (p-q)² is of the form a[x² - (4mn)x - (m²-n²)²] for a≠0.

This is about number theory.

That's right, that's the same solution I showed it. [a ≡ b (mod c) means that c | a-b]

If n ≡ 0 (mod 6), n³ ≡ 0 (mod 6)

If n ≡ 1 (mod 6), n³ ≡ 1 (mod 6)

If n ≡ 2 (mod 6), n³ ≡ 8 ≡ 2 (mod 6)

If n ≡ 3 (mod 6), n³ ≡ 27 ≡ 3 (mod 6)

If n ≡ 4 (mod 6), n³ ≡ 64 ≡ 4 (mod 6)

If n ≡ 5 (mod 6), n³ ≡ 125 ≡ 5 (mod 6)

This is about number theory.

3^(2n)+7 = (9^n - 1)+8 = (9^n - 1^n)+8 = (9-1)(9^(n-1) + 9^(n-2) +...+ 1)+8 = 8(9^(n-1) + 9^(n-2) +...+1 +1)

So 8 | 3^(2n)+7 for all n ∈ N∪{0}

It would be incredibly difficult to type the solution on mobile, so I’ll give it when I get back to my house

(a+b+c+d)(1/a​+4/b​+9/c​+16/d​)​≥(sqrt(a)sqrt(1/a)​​+sqrt(b)(sqrt(4/b)​​+sqrt(c)(sqrt(9/c)​​+sqrt(d)(sqrt(16/d)​​)^2≥(1​+4​+9​+16​)2≥(1+2+3+4)2≥100

Since a+b+c+d=1, we can see that 100 is a lower bound for 1/a+4/b+9/c+16/d. The lower bound is the minimum only when we have each equality, which is when:

(sqrt(a)/sqrt(1/a))=(sqrt(b)/sqrt(4/b))=(sqrt(c)/sqrt(9/c))=(sqrt(d)/sqrt(16/d))​, that is when a/1=b/2=c/3=d/4. Let r=a/1=b/2=c/3=d/4. Then by the condition a+b+c+d=1, we have r+2r+3r+4r=1, and r=1/10 which gives us a=1/10, b=2/10, c=3/10, and d=4/10.

When you plug these in, the value is 100, which tells us that the minimum value is 100.

PROBLEM: 7

Let [A,B,C are positive real number]ᴱᴰᴵᵀ that A², B², C² are arithmetic sequence, prove that A+B, A+C, B+C are harmonic sequence.