n, n+1 - two consecutive integers
n(n + 1) = 50 <em>use distributive property</em>
n² + n = 50 <em>subtract 50 from both sides</em>
n² + n - 50 = 0
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ax² + bx + c =0
if b² - 4ac > 0 then we have two solutions:
[-b - √(b² - 4ac)]/2a and [-b - √(b² + 4ac)]/2a
if b² - 4ac = 0 then we have one solution -b/2a
if b² - 4ac < 0 then no real solution
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n² + n - 50 = 0
a = 1, b = 1, c = -50
b² - 4ac = 1² - 4(1)(-50) = 1 + 200 = 201 > 0 → two solutions
√(b² - 4ac) = √(201) - it's the irrational number
Answer: There are no two consecutive integers whose product is 50.
<h2>
The value of r = 
</h2>
Step-by-step explanation:
We have,
U = F(r + B)
To find, the value of r = ?
∴ U = F(r + B)
Dividing both sides by F, we get
⇒ 
= r + B
⇒ r + B =
⇒ r = - B
⇒ r =
∴ The value of r =
You can do synthetic division twice immediately (or once if you know how to handle x^2 - 25 all at once. I'll use 2 divisions. x^2 - 25 is (x + 5)(x - 5). Both of them are zeros to the given equation which means that x = +5 and
x = - 5 are both zeros of the given quartic.
5 1 -2 -33 50 200
5 15 -90 -200
=====================================
1 3 -18 -40 0
What you have now is x^3 + 3x^2 - 18x - 40 = 0
Do another synthetic division
-5 1 3 -18 -40
-5 10 40
============================
1 -2 -8 0
The result is
x^2 - 2x - 8 = 0 which factors.
(x + 2)(x - 4) = 0
Answer
The factors are (x + 2)(x - 4)(x + 5)(x - 5) = 0
Long division
x^2 + 0 + 25 ||x^4 - 2x^3 - 33x^2 + 50x + 200 ||x^2 - 2x - 8
x^4 + 0 + 25x^2
==================
-2x^3 -8x^2 + 50x
-2x^3 + 0 + 50x
=================
-8x^2 + 0 +200
-8x^2 + 0 + 200
==================
0
Can be written as 24% or 24/100
