Algebra discussion

Put the theorems, problems, homework or anything else related to Algebra here. We're here to help if we can.

After trying to do for 2 hours. I finally prove it!

For example : when I want to find x such that p(x)=x²-x+41 is prime. I can choose any integer k, where p(k)+k = x then p(x) is composite.

Choose k=1 ; p(k)+k = 41+1 = 43 so p(43) is composite.

Choose k=2 ; p(k)+k = 43+2 = 45 so p(45) is composite. etc.

I think this is an algebra question—it was on one of the math state tests for our schools club.

Given the natural numbers k, m, and n where k is greater than or equal to m, k is greater than or equal to n, and m does not equal n, what is the sum of m and n if C(k,m)=C(k,n)? 

Yes, this is an algebra problem. You can write the function C in factorial form and format the equation. However, this problem can use combinatorial proof which intuitive.

There are k items. When you select m items, you can do in C(k,m) ways. Or if selected n items, the number of ways to choose is the same. Is it possible that m ≠ n?

The answer is yes! When m≠n imagine that if you go from selecting m items to opting out m items, the number of methods obviously remains the same. And that is equivalent to the selection of k-m items, so the value of n that makes the equation true is k-m.

So m+n = m+(k-m) = k

Problem 1:

How many integers b are there in total? So that the equation 4x²+bx+15=0 to have no solution for x in real number system.

The equation ax²+bx+c=0 have a solution x = [-b ± root_/b²-4ac] / 2a

So, x is not a real number when b²-4ac < 0.

Correct!

Problem 2:

Find the largest area possible. of a rectangle with a perimeter of 100 cm.

You have now shown that the area of ​​a rectangle is 50L-L². So we will find its maximum value by taking the form of absolute squares.

-L²+50L = -L²+2(L)(25)-25² +25² = -(L²-50L+25²)+25² = -(L-25)²+25² = 625 - (L-25)²

It can be seen that the maximum area value is 625 cm², occurring when L=25 cm. [L=B=25 is square].

x² - y² = (x+y)(x-y)

This is an easy one. Express tan(2x) all in terms of sin(x) with no numbers. If a number is left, turn it into a sine function.
On another note, over this summer, I’ll be doing intensive maths studies to improve my AMC and AIME score, so I expect my math skills to improve drastically.  Pure mathematics is gonna be my major in college.

tan(2x) = sin(2x)/cos(2x)

Consider: sin(2x) = 2sinx cosx = 2sinx(1-sin²x)⁰'⁵

cos(2x) = (1-sin²(2x))⁰'⁵ = [1-(2sinx(1-sin²x)⁰'⁵)²]⁰'⁵ = [1-4sin²x(1-sin²x)]⁰'⁵ = [4sin⁴x-4sin²x+1]⁰'⁵ = [(2sin²x-1)²]⁰'⁵ = ±(2sin²x-1)

tan(2x) = ± 2sinx(1-sin²x)⁰'⁵ / (2sin²x-1)

Problem 3:

There are two real numbers whose difference and product are equal to 5, find the sum of the two numbers.

Correct! Actually, the answer to this problem can be both positive and negative. I should define x,y as positive real numbers.

(x+y)² = (x-y)² + 4xy

(x+y)² = 5² + 4(5) = 45

x+y = (±) 3×(root_/5) ~ 6.7

Problem 4:

Let 0 < x < 1 prove that 0 < x¹⁰-x²⁰ ≤ 0.25

Yes, calculus can help find the maximum as well. Notice that the vertex of the parabola has a slope of 0.

This inequality may be difficult to see. It's best to look at x¹⁰ terms, so x²⁰ = (x¹⁰)².

Prove L.S. since 0 < x < 1 makes x¹⁰ > 0 and 1 > x¹⁰ that's 1-x¹⁰ > 0. So (x¹⁰)(1-x¹⁰) = x¹⁰-x²⁰ > 0

R.S. for all x∈R we can say that 0 ≤ (x¹⁰-0.5)² = x²⁰-2(0.5)(x¹⁰)+0.25 will get 0 ≤ x²⁰-x¹⁰+0.25 so -0.25 ≤ x²⁰-x¹⁰ => 0.25 ≤ x¹⁰-x²⁰

So 0 < x¹⁰-x²⁰ ≤ 0.25 for all 0 < x < 1.

The inequality is an equation only if x¹⁰-0.5 = 0, that is, x = root_10 (0.5) ~ 0.933

Problem 5:

Find the value of c, such that the line y=2x+c is tangent to the parabola y=x².