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For which values of the number a (identity)?
For which values of the number a
xa + 2a = x + 2a^2 is an identity
The answer is going to be a=1. How can I get that answer? Can someone show me step by step because I want to learn. Thank you.
5 Answers
- ?Lv 75 months ago
ax+2a=x+2a^2
=>
x(a-1)=2a(a-1)
=>
(x-2a)(a-1)=0
So, if a=1, then
(x-2)*0=0
=>
0=0 is an identity, no matter
what x is.
- Ray SLv 75 months ago
xa + 2a = x + 2a² ← We're asked to determine the value of "a".
So, we want to solve for "a", not x
��� 0 = x + 2a² - xa - 2a ← Got everything on one side and equal to 0.
Now, begin factoring.
0 = x - xa + 2a² - 2a ← Rearranged the terms
0 = x(1 - a) + 2a(a - 1) ← Notice that (1-a) and (a-1)are opposites ...
i.e. -(1-a)=(a-1) and -(a-1)=(1-a)
So, +2a(a - 1) = -2a(1-a)
0 = x(1 - a) - 2a(1 - a) ← Now, factor out (1 - a)
0 = (x - 2a) (1 - a ) ← Now, apply the Zero Product Property to solve for "a".
n/a , 1- a = 0 Since x can be any number, we can't get a definite
ANSWER → a = 1 value for "a" by considering factor (x-2a).
But, from the factor (1-a), we get that a=1
Checking
xa + 2a = x + 2a²
x*1 + 2*1 = x + 2*1²
x + 2 = x + 2 ✔
- PopeLv 75 months ago
xa + 2a = x + 2a²
2a² + (-x - 2)a + x = 0
This is a quadratic equation in a. Apply the quadratic formula.
a = {x + 2 ± √[(-x - 2)² - 8x]} / 4
a = {x + 2 ± √[(x - 2)²]} / 4
a = [x + 2 ± (x - 2)] / 4
a = x/2 or a = 1
That first solution, a = x/2, is at least a little shaky, because a would not be a constant. Could we really call that a value of a when x itself is unknown?
The solution a = 1 satisfies the condition without a doubt.
- Wayne DeguManLv 75 months ago
Equating coefficients of x we have:
xa = x
so, a = 1
Also, equating constant terms we have:
2a = 2a²
Again, a = 1
Then, with a = 1 we have:
x + 2 ≡ x + 2
:)>
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- la consoleLv 75 months ago
xa + 2a = x + 2a²
xa - x = 2a² - 2a
x.(a - 1) = 2a² - 2a
x = (2a² - 2a)/(a - 1)
x = 2a.(a - 1)/(a - 1)
x = 2a
a = x/2