## anonymous one year ago .

1. anonymous

I'm not sure sorry.

2. anonymous

Np

3. princeharryyy

x^2 + 4x +x +4 =0 x(x+4) + 1(x+4) =0 (x+1)(x+4)=0 =>x=-1,-4

4. anonymous

How do we define radicand?

5. princeharryyy

yeah

6. anonymous

Radicand is the discriminate or b^2 - 4ac and @princeharryyy please do not solely give answers

7. anonymous

So If the discriminant is positive, there are two real roots. If it is zero, there is one real root (a double root). And if the discriminant is negative, there are two imaginary roots and no real roots

8. campbell_st

the radicand is the discriminant so you need to use for a quadratic $ax^2 + bx + c = 0$ the discriminant is $\Delta = b^2 - 4ac$

9. princeharryyy

tnks, i'm not in mood as well @whatdoesthismean by the way discriminate means (The value you want to take the root of)

10. princeharryyy

11. anonymous

By plugging a b and c, you can find the answer to A as it is asking for the DESCRIPTION of the solutions, not the solutions itself

12. campbell_st

so for the other parts.. Part B the quadratic can be factored Pact C use the general quadratic formula

13. anonymous

as i said before, the descriptions are as follows'\; So If the discriminant is positive, there are two real roots. If it is zero, there is one real root (a double root). And if the discriminant is negative, there are two imaginary roots and no real roots

14. anonymous

Agreed @campbell_st

15. campbell_st

and when you do part A you have a = 1, b = 5 and c = 4 and if the discriminant is positive there are 2 options (a) the discriminant is a sqiare number, 1, 4, 9, 16.... then the roots are real, rational and unequal (b) is the discriminant is not a square number e.g. 3, 7, 10, 15, etc the roots are real, irrational and unequal

16. anonymous

No, youre supposed to try and use the information and solve it

17. campbell_st

ok... you need $\Delta = b^2 - 4\times a \times c$ the equation have a = 1, b = 5 and c = 4 plug them in post the answer you get..

18. campbell_st

well $\Delta = 5^2 - 4 \times 1 \times 4$ what is the value of the calculation

19. campbell_st

ummm can you recheck it $\Delta = 25 - 16$

20. campbell_st

ummmm I think $\Delta = 25 - 16~~~means~~~~\Delta = 9$ so based on the solution being >0 and a square number, what type of solutions do you have...?