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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I'm not sure sorry.
Np
x^2 + 4x +x +4 =0 x(x+4) + 1(x+4) =0 (x+1)(x+4)=0 =>x=-1,-4

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How do we define radicand?
yeah
Radicand is the discriminate or b^2 - 4ac and @princeharryyy please do not solely give answers
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
the radicand is the discriminant so you need to use for a quadratic \[ax^2 + bx + c = 0\] the discriminant is \[\Delta = b^2 - 4ac\]
tnks, i'm not in mood as well @whatdoesthismean by the way discriminate means (The value you want to take the root of)
radical*
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
so for the other parts.. Part B the quadratic can be factored Pact C use the general quadratic formula
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
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
No, youre supposed to try and use the information and solve it
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..
well \[\Delta = 5^2 - 4 \times 1 \times 4\] what is the value of the calculation
ummm can you recheck it \[\Delta = 25 - 16\]
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...?

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