Question

need help with the attached problem.

Answer 1

ax^2+bx+c=0
when b^2-4ac>=0,
the equation has real roots

Answer 2

so is a=1+2k, b=-10+k and c=-2?

Answer 3

(10/2)^2-(k-2)(1+2k)=-2k^2-3k+27 (1)
solution of the equation (1) are 3, -9/2
(1)>=0
-9/2<=k<=3

Answer 4

no
a=1+2k
b=-10
c=k-2

Answer 5

b^2-4ac>=0
or
(b/2)^2-ac>=0

Answer 6

my text gives an different answer then yours?

Answer 7

oh sorry

(10/2)^2-(k-2)(1+2k)=-2k^2+3k+27 (1)
solution of the equation (1) are -3, --9/2
(1)>=0
-3<=k<=9/2

Answer 8

my text gives the answer, k=4.08 and k=-2.58

Answer 9

are you sure

Answer 10

\[ (1+2k)x^2-10x +(k-2)=0 \] k real
will have real roots when its discriminant \(b^2-4ac ≥0\)
See http://www.mathwarehouse.com/quadratic/discriminant-in-quadratic-equation.php

Here a (coefficient of x^2 term) is (1+2k)
b is the coefficient of the x term, -10
c is the constant term (k-2)
the discriminant is
100-4*(1+2k)(k-2) ≥ 0
which simplifies to
\[ -2k^2+3k+27≥0 \]
we can say
\[ y= -2k^2+3k+27 \]
and we want y≥0
y is the equation of a negative parabola (makes a "frown")