solve the following equations. 3x^2-5x+2=0. x^-x=20. x-1=6/x. x^4-5x^2+4=0. 9x^4-13x^2+4=0

Question

i really need help (again).... i understand how to factorize now and even solve but i dont know how to solve an equation with a number infront of the first variable or the steps to solve the rest. can someone please help me?

Answer 1

3x²-5x+2=0

(3x-2)(x-1)=0

3x-2=0

3x=2

x=2/3

x-1=0

x=1

 

x²-x=20

x²-x-20=0

(x-5)(x+4)=0

x-5=0

x=5

x+4=0

x=-4

 

x-1=6/x

x²-x=6

x²-x-6=0

(x-3)(x+2)=0

x-3=0

x=3

x+2=0

x=-2

 

x⁴-5x²+4=0

let y=x²

y²-5y+4=0

(y-4)(y-1)=0

y-4=0

y=4, so x²=4 and x=+2

y-1=0

y=1, so x²=1 and x=+1

Answer 2

All these problems are variations on the theme of quadratic equations. So if you know how to solve the general form of a quadratic equation x²+px+q=0 then you can solve the ones above if can convert them to this format.

 

In the first case you would divide the given equation by the coefficient 3 of x², giving x²-5/3*x+2/3=0. So p = -5/3, q = 2/3.

 

In the second (I take it you forgot the exponent 2 in the first term) you subtract 20 on both sides, giving x²-5x-20=0.

So p = -5, q = -20.

 

In the third case you have x appear in the denominator in a term on the right side, so you have to do steps: multiply the equation by and then subtract 6. This yields x²-x-6=0, thus p = -1 and q = -6.

 

And in the last case you have x appear in the second and fourth power; here you substitute y=x², giving you y²-13y+4=0, which you can solve for y. Once you have the two solution for y, you take the square root of each and end up with a total of four solutions for x.