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I need some help with 2 questions.

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

[tex]7{x}^{2} - 14 = 0[/tex]
b) the equation
[tex](k + 3) {x}^{2} + 6x + k = 5[/tex]
, where k is a constant, has 1 distinct real solution.

i) by letting the discriminant = 0, show that k satisfies
[tex] {k}^{2} - 2k - 24 = 0[/tex]

ii) hence find the possible values of k.​

Respuesta :

Answer:

Step-by-step 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

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