Thursday

November 27, 2014

November 27, 2014

Posted by **wi** on Monday, May 21, 2007 at 2:13pm.

I am stuck on this question. y=x^2+4x and y^2= 25x^2+20x+3

how many of the solutions are real? and how many of the real roots are positive and how many are negative?

thank you for any suggestions

I suggest you solve for the roots.

y=x(x+4) At y=0, what are the roots?

Y= (5x+3)(5x+1)

**Answer this Question**

**Related Questions**

algebra - determine the nature of the solutions of the equation, one real ...

Algebra II - Which describes the number and type of roots of the equation x^2 -...

Algebra - in x^2+14=0, how many solutions do we have? A)2 real solutions B)1 ...

algebra - Determine the value(s) of k for which x^2+(k-2)x-2k=0 has equal and ...

maths2 - Use the discriminant to determine the number of real roots the equation...

mathematics - Use the discriminant to determine the number of real roots the ...

Math - I am given 4 choices for the problem the square root of 8x -2 = the ...

algebra - determine the nature of the solutions of the equation. Two real ...

Algebra - How many real number solutions does the equation have? y=-4x^2+7x-8 A...

Algebra - I am stuck... find the real solutions of the equation x to the 4th ...