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luda_lava [24]
3 years ago
15

Hi. Please help me with these questions See image for question.Answer no 6 and 7

Mathematics
1 answer:
Gnom [1K]3 years ago
7 0

Answer:

  • 6. See solution
  • 7. k = 2, k = -6

Step-by-step explanation:

6.

<u>Given equation:</u>

  • 2(x + 2)² + p(x + 1) = 0
  • The sum of the roots is q1 and the product of the roots is q2

Need to show that q1+q2 = 0

<h3>Solution</h3>

<u>Bringing the equation into standard form of ax² + bx + c = 0:</u>

  • 2(x + 2)² + p(x + 1) = 0
  • 2x² + 8x + 8 + px + p = 0
  • 2x² + (p + 8)x + (p + 8) = 0

<u>Sum of the roots: </u>

  • q1 = - b/a = -(p + 8)/2

<u>Product of the roots:</u>

  • q2 = c/a = (p + 8)/2

<u>We see that q1 and q2 are opposite numbers, therefore their sum equals zero:</u>

  • q1 + q2 = -(p + 8)/2 +  (p + 8)/2 = 0

=============================================

7.

<u>Given quadratic equation:</u>

  • x² - (k + 2)x + 4 = 0
  • It has equal roots

Need to find the possible values of k

<h3>Solution</h3>

<u>When the quadratic equation has equal roots, then its discriminant is equal to zero:</u>

  • 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
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