x^2 + 4x + c = (x + a)^2 + b
x^2 +4x +c = x^2 + 2ax + a^2 + b
4x + c = 2ax + a^2 + b
Since the coefficients of the x^1 and the x^0 must be equal (as x varies), 4x=2ax and thus 4=2a or a=2, and c=a^2+b or c=4+b
Given also that the equation x^2 + 4x + c = 0 has unequal real roots:
iii. find the range of possible values of c.
Using the discriminant (b^2-4ac, the part of the quadratic formula that we take the square root of), we can determine what values of c give real roots.
If the discriminant is 0, there is one repeated real root, because the solutions given by the quadratic equation are of the form (-b +/- sqrt (discriminant))/2a and if the discriminant is zero the solutions are -b/2a. If the discriminant is real and greater than 0, there are two real roots. If the discriminant is negative, the roots are imaginary because the square root of a negative number is a complex number. You are then solving for b^2-4ac > 0, I'll leave that to you.
More when I can get back on...
Jeff
x^2 +4x +c = x^2 + 2ax + a^2 + b
4x + c = 2ax + a^2 + b
Since the coefficients of the x^1 and the x^0 must be equal (as x varies), 4x=2ax and thus 4=2a or a=2, and c=a^2+b or c=4+b
Given also that the equation x^2 + 4x + c = 0 has unequal real roots:
iii. find the range of possible values of c.
Using the discriminant (b^2-4ac, the part of the quadratic formula that we take the square root of), we can determine what values of c give real roots.
If the discriminant is 0, there is one repeated real root, because the solutions given by the quadratic equation are of the form (-b +/- sqrt (discriminant))/2a and if the discriminant is zero the solutions are -b/2a. If the discriminant is real and greater than 0, there are two real roots. If the discriminant is negative, the roots are imaginary because the square root of a negative number is a complex number. You are then solving for b^2-4ac > 0, I'll leave that to you.
More when I can get back on...
Jeff