linto asked in Science & MathematicsMathematics · 2 months ago

# 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.

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• ?
Lv 7
2 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 S
Lv 7
2 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     ✔

• Pope
Lv 7
2 months ago

xa + 2a = x + 2a²

2a² + (-x - 2)a + x = 0

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.

• 2 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

:)>

• 2 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