Answer:
A Quadratic Equation can have upto 2 roots maximum. So,if one of the roots is a Real number, there are following two possibilities:
1) The other root is also a real number, but a different number
2) Its a repeated root, so the other root is the same number.
The other root cannot be a complex number as its not possible for one root to be real and other to be complex. Either no root will be complex or both will be complex roots.
Following are 3 possibilities for the roots of a quadratic equation:
- 2 Real and Distinct roots
- 2 Real and Equal roots
- 2 Complex roots
Answer:
(3x^2 + 2)(x + 4).
Step-by-step explanation:
3x3 + 12x2 + 2x + 8
3x^2(x + 4) + 2(x + 4) The x + 4 is common to the 2 groups so we have:
(3x^2 + 2)(x + 4).
Ok so for starters you want to choose an equation and solve for a variable.
So, I am going to choose x from the first equation.
Add y to both sides and you get x=11+y
Next, substitute 11+y for x in the other equation so you get...
2(11+y) +10y=-6
Next distribute the 2 throug the 11 and the y
22+2y+10y=-6
12y=-28
y=-28/12
reduce this fraction to make this easier.
y=-7/3
Now plug in why to either of the equations to find x
x-(-7/3)=11
x+7/3=11
x=11-(7/3)
x=(33/3)-(7/3)
x=26/3
so x = 26/3 and y = -7/3
you can also check to see if this is correct by substituting each of these values into the equations.
Answer:
x = 0.5
Step-by-step explanation:
Given
2(x + 1.25) = 3.5 ← divide both sides by 2
x + 1.25 = 1.75 ( subtract 1.25 from both sides )
x = 0.5
