Answer:
- 6. See solution
- 7. k = 2, k = -6
Step-by-step explanation:
6.
Given equation:
- The sum of the roots is q1 and the product of the roots is q2
Need to show that q1+q2 = 0
Solution
Bringing the equation into standard form of ax² + bx + c = 0:
- 2(x + 2)² + p(x + 1) = 0
- 2x² + 8x + 8 + px + p = 0
- 2x² + (p + 8)x + (p + 8) = 0
Sum of the roots:
Product of the roots:
We see that q1 and q2 are opposite numbers, therefore their sum equals zero:
- q1 + q2 = -(p + 8)/2 + (p + 8)/2 = 0
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7.
Given quadratic equation:
Need to find the possible values of k
Solution
When the quadratic equation has equal roots, then its discriminant is equal to zero:
- D = 0
- √b² - 4ac = 0
- √(-k -2)² - 4*1*4 = 0
- √k² + 4k + 4 - 16 = 0
- √k² + 4k - 12 = 0
- k² + 4k - 12 = 0
- k = {-4 ± √4² -4*1*(-12)}/2
- k = (-4 ± √16 + 48)/2
- k = (-4±√64)/2
- k = -2 ± 4
- k = 2, k = -6
