Solve by factorization method :::::

- x
^{2}+ 3x - (a^{2}+ a - 2) = 0 - a
^{2}b^{2}x^{2}+ b^{2}x - a^{2}x - 1 = 0 - x
^{2}+ x - (a + 1)(a + 2) = 0 - x
^{2}+ ({a/a +b} + {a + b/a}) - 1/a+b+x = 1/a + 1/b + 1/x

Dear Student!

Here is the answer to your question.

1.

⇒

*x*^{2}+ 3*x*– (*a*^{2}+*a*– 2) = 0⇒

*x*^{2}+ 3*x*– (*a*+ 2) (*a*– 1) = 0⇒

*x*^{2}+ [(*a*+ 2)] – (*a*– 1)]*x*– (*a*+ 2) (*a*– 1) = 0⇒ –

*ab*=*x*^{2}+*ax*+*bx*⇒

*x*^{2}+*ax + bx + ab = 0*⇒

*x*(*x*+*a*) +*b*(*x*+*a*) = 0⇒ (

*x*+*b*) (*x*+*a*) = 0⇒

*x + b*= 0 or*x*+*a*= 0⇒

*x*= –*b*or*x*= –*a*Thus, the solutions of given quadratic equation are –

*a*, –*b*.Cheers!

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