# Find the set of values of k?

Find the set of values of k for which the line y = 2x + k cuts the curve y = x2 + kx + 5 at two

distinct points.

PLease include simple explanations on each steps.

### 3 Answers

- la consoleLv 74 years agoBest Answer
y = 2x + k ← this is a line

y = x² + kx + 5 ← this is a curve

y = y → 2 points → 2 solutions for x

x² + kx + 5 = 2x + k

x² + kx + 5 - 2x - k = 0

x² + x.(k - 2) + (5 - k) = 0

Polynomial like: ax² + bx + c, where:

a = 1

b = (k - 2)

c = (5 - k)

Δ = b² - 4ac (discriminant)

Δ = (k - 2)² - 4.(5 - k)

Δ = k² - 4k + 4 - 20 + 4k

Δ = k² - 16

Δ = (k + 4).(k - 4) → 2 solutions for x → Δ > 0 → the roots are (- 4), (4) → then you make a table

k_____-∞____- 4____4____+∞

(k + 4)_____-__0__+____+

(k - 4)______-____-__0__+

Δ sign_____+__II_-__0__+

…and you can see when the sign is > 0

→ k Є ] -∞ ; - 4 [ U ] 4 ; +∞ [

x₁ = (- b - √Δ)/2a → y₁ = 2x₁ + k

x₂ = (- b + √Δ)/2a → y₂ = 2x₂ + k

- Raj KLv 74 years ago
y = 2x + k cuts the curve y = x2 + kx + 5

→ 2x + k=x2 + kx + 5

or x²+(k−2)x+5−k=0

Hence a=1, b=k−2 and c=5−k

Discriminant D=b²−4ac=(k−2)²−4×1×(5−k) > 0

→k²−4k+4−20+4k> 0

i.e. →k²−16> 0

k²>16 or either k > 4 or k < −4

Hence set of values is (−∞,−4)U(4,∞ )

- cidyahLv 74 years ago
y=2x+k

y= x^2+kx+5

x^2+kx+5 = 2x+k

x^2+kx-2x+5-k=0

x^2+x(k-2)+(5-k)= 0

ax^2+bx+c = 0

a=1

b=k-2

c=5-k

discriminant = b^2-4ac > 0 for real roots

= (k-2)^2 - 4(5-k)

= k^2-4k+4-20+4k

= k^2-16 > 0

solve k^2-16 =0

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

k=-4,4

Consider the intervals:

(-∞ ,-4),(-4,4),(4, ∞)

choose any one point from each interval and examine the sign of k^2-16

(-∞ ,-4): choose k=-5

k^2-16 = 9 > 0 (a solution)

(-4,4) : choose k = 0

k^2-16 = -16 < 0 (solution does not hold)

(4, ∞): choose k= 5

k^2-16 = 9 > 0 (solution)

Therefore: k < -4 and k > 4 are solutions

(-∞ ,-4) U (4, ∞)