Answer:
340,000
Step-by-step explanation:
Consecutive odd integers are 2 apart
they are x and x+2
the product (x times (x+2)) is 1 less than 4 times their sum
x(x+2)=-1+4(x+x+2)


![x^2+2x=8x+7 minus (8x+7) from both sides [tex]x^2-6x-7=0](https://tex.z-dn.net/?f=x%5E2%2B2x%3D8x%2B7%20minus%20%288x%2B7%29%20from%20both%20sides%20%5Btex%5Dx%5E2-6x-7%3D0)
factor
what numbers multliply to get -7 and add to get -6?
-7 and 1
(x-7)(x+1)=0
set equal to 0
x-7=0
x=7
x+1=0
x=-1
so the x can be 7 or -1
the other number (x+2) can be 7 or 1
test each
7 and 9
7(9)=-1+4(7+9)
63=-1+4(16)
63=-1+64
63=63
true
-1 and 1
-1(1)=-1+4(-1+1)
-1=-1+4(0)
-1=-1+0
-1=-1
true
the 2 integers can be 7 and 9 or -1 and 1 (both pairs of numbers work)
I don’t really know but try a calculator
The answer is a because its 125 divided by 4
Answer:
- The required value of q is 35.
Step-by-step explanation:
Let α and β are the zeros of quadratic equation, x^2−12x+q=0.
- It is given that difference between the roots of the quadratic equation x^2−12x+q=0 is 2.
Equation : α - β = 2
Equation : α + β = 12
Equation : αβ = q
We have to create an algebraic expression.
(a+b)² = (a-b)² + 4ab
(12)² = (2)² + 4q
144 = 4 + 4q
144 - 4 =4q
140=4q
q = 140/4
q = 35
Therefore, the required value of q is 35.
<u>Some information about zeroes of quadratic equation. </u>
- Sum of zeroes = -b/a
- Product of Zeroes = c/a