Question

Find the values of for which x2 + 6x + k = 0 has two real solutions.

Find the set of values of x for which:
a. 3x – 5 < x + 8 and 5x > x – 8
b. x – 5 > 1 – x and 15 – 3x > 5 + 2x

Answer 1

Given quadratic equation can be written as,

x

2

−(2m)x+(8m−15)=0

Now for equal roots, Discriminant =0

⇒b

2

−4ac=0

⇒(−2m)

2

−4(8m−15)=0

⇒m

2

−8m+15=0

⇒(m−3)(m−5)=0, factor

⇒m=3,5

Answer 2

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