Question

ax²+a = a² + x find the root of this equation​

Answer 1

✏Solution :-

ax² +a = a² +x

=> ax² - x +(a+a²)=0

=> x = [ 1 ± √ {1² -4×a×(a+a²)}]/2a

=>x = [ 1± √ { 1 - 4a² - 4a³} ]/2a

=> x = [ 1±√(1-4a³-4a²)]/2a

take first +ve sign and negative ,

✏you can find your roots .

Answer 2

Answer:

The given equation is:

ax^2 + a = a^2 + x

ie. ax^2 + a - a^2 - x = 0

=> ax^2 - a^2 + a - x = 0

=> ax(x - a) - (x - a) = 0

=> (x - a)(ax - 1) = 0

Now , if (x - a) = 0

then , x = a.

And, if ( ax - 1) = 0

then, x = 1/a.

Hence, the required roots of the given equation are: a and 1/a.