# Math Help - Quadratic equation

1)For what value of a,quadratic equation x^2(a^2-5a+6)+x(a^2-3a+2)+a^2-4 =0 will have more than two roots.

2)if the equation x^2(a^2+a-2)+x(a^2+m-3)+(a^2-1)=0 is satisfied by more than two values of x. find the values of a and m.

MENTION ALL THE STEPS .

1)For what value of a,quadratic equation x^2(a^2-5a+6)+x(a^2-3a+2)+a^2-4 =0 will have more than two values of x.

2)if the equation x^2(a^2+a-2)+x(a^2+m-3)+(a^2-1)=0 is satisfied by more than two values of x. find the values of a and m.

MENTION ALL THE STEPS .
I don't get what you mean by "more than two values of x". By my understanding quadratics only have two roots (ie: cross the x axis). These can be real and different, real and equal or complex

1)For what value of a,quadratic equation x^2(a^2-5a+6)+x(a^2-3a+2)+a^2-4 =0 will have more than two roots.
Normally, a quadatic (the coefficient of $x^2$ not equal to 0) will have two roots and a linear equation (the coefficient of $x^2$ equal to 0 but the coefficient of x not equal to 0) will have 1. This will have an infinite number of solutions only if $a^2- 5a+ 6= 0$, $a^2- 3a+ 2= 0$ and $a^2- 4= 0$. What is a common factor of all of those?

2)if the equation x^2(a^2+a-2)+x(a^2+m-3)+(a^2-1)=0 is satisfied by more than two values of x. find the values of a and m.
Same thing: for what values of m do $a^2+ a- 2$, $a^2+ m- 2= 0$, and $a^2- a$ have a common factor and what is that common factor?

MENTION ALL THE STEPS .
Since this is your exercise, why don't you at least try before asking someone to do it for you?

5. The only way I can think of for you to have more than 2 $x$ intercepts in either case is to have $a$ be a function of $x$...

Apparently you have NOT tried doing exactly what I told you to do: "This will have an infinite number of solutions only if $a^2- 5a+ 6= 0
, $a^2- 3a+ 2= 0$ and $a^2- 4= 0$. What is a common factor of all of those?" For what value of a will all of those be 0?
The only way I can think of for you to have more than 2 $x$ intercepts in either case is to have $a$ be a function of $x$...