Question

I need some help with 2 questions.

a) using factorisation, solve for x in the following equation:


`7{x}^(2) - 14 = 0`
b) the equation

`(k + 3) {x}^(2) + 6x + k = 5`
, where k is a constant, has 1 distinct real solution.

i) by letting the discriminant = 0, show that k satisfies

`{k}^(2) - 2k - 24 = 0`

ii) hence find the possible values of k.​

Answer 1

Explanation:

a.

7x²-14=0

x²-2=0

(x+√2)(x-√2)=0

x=±√2

b.

disc.=6²-4(k+3)(k-5)=36-4(k²-5k+3k-15)

=-4(k²-2k-15)+36

=-4k²+8k+60+36

=-4k²+8k+96

=-4(k²-2k-24)

as it has 1 real solution.

so -4(k²-2k-24)=0

k²-2k-24=0

k²-6k+4k-24=0

k(k-6)+4(k-6)=0

(k-6)(k+4)=0

k=6,-4