Question

if a and b are distinct positive integers and the quadratic equations (a-1)x^2 +(a^2+2)x +(a^2+2a) = 0 and ( b-1) x^2 -(b^2+2) x +( b^2+2b) = 0 have a common root then ab =​

Answer 1

Answer:

x-6/x=3 is an quadratic equations in x and if yes then give explanation please...

Answer 2

The value of ab = (a + b + 2).

Quadratic equation

Quadratic equations are the polynomial equations of degree 2 in one variable of type $f(x) = ax^{2} + bx + c$ where a, b, c, ∈ R and a ≠ 0.

Given:

The two quadratic equation are,

$(a-1)x^{2} -(a^{2} +2)x+(a^{2}+2a )=0$

$(b-1)x^{2} -(b^{2}+2 )x+(b^{2} +2b)=0$

Explanation:

Equating the coefficients of the variable,

$\frac{a-1}{b-1}=\frac{-(a^{2}+2 )}{-b^{2}+2 } =\frac{a^{2}+2a }{b^{2}+2b }$

Now, comparing first two equation we get,

$\frac{a-1}{b-1}=\frac{a^{2}+2 }{b^{2} +2}$

Solving the equation we get,

$(a-b)[-ab+a+b+2]=0$

As, (a-b)≠0, (a,b are two distinct positive integers)

[-ab + a + b + 2] = 0

ab = a + b + 2

Thus the value of ab = a + b + 2.

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