I need some help with 2 questions.

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

b) the equation

, where k is a constant, has 1 distinct real solution.

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

ii) hence find the possible values of k.

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

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation:

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.