Q5. If -3 and 2 are roots of the quadratic equation x²-(p+2)x-q=0, then the values of p and q are: * 1 point 1) 3, -6<

Question

Q5. If -3 and 2 are roots of the quadratic equation x²-(p+2)x-q=0, then the values of p and q are: *

1 point

1) 3, -6

2) -3, 6

3) -3, -6

4) 3, 6

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Savannah 4 days 2021-10-10T23:27:55+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-10T23:29:18+00:00

    Step-by-step explanation:

    Given –

    • -3 and 2 are roots of quadratic equation x² – (p + 2)x – q = 0

    To Find –

    • Value of p and q

    Now,

    p(x) = x² – (p + 2)x – q = 0

    Then,

    → p(-3) = (-3)² – (p + 2)-3 – q = 0

    → 9 – (-3p – 6) – q = 0

    → 9 + 3p + 6 – q = 0

    → 3p – q = -15 …. (i)

    And

    p(2) = (2)² – (p + 2)2 – q = 0

    → 4 – (2p + 4) – q = 0

    → 4 – 2p – 4 – q = 0

    → -2p – q = 0 ….. (ii)

    Now,

    From (i) and (ii), we get :-

    → 3p – q = -15

    -2p – q = 0

    (+) (+)

    ____________

    → 5p = -15

    → p = -3

    Now,

    Substituting the value of p on -2p – q = 0, we get :-

    → -2(-3) – q = 0

    → 6 – q = 0

    → q = 6

    Hence,

    The value of p is -3 and q is 6

    Therefore,

    Option 2 is correct.

    0
    2021-10-10T23:29:34+00:00

    Answer:

    Step-by-step explanation:

    p and q are the roots of the equation x2 + px + q = 0, then

    p = 1 and q = -2

    p = 0 and q = 1

    p = -2 and q = 0

    p = -2 and q = 1

    Correct Option: A

    Since, p and q are the roots of the equation x2 + px + q = 0

    Then, p + q = – p

    and pq = q

    Now, pq = q

    ⇒ p = 1

    Putting the value of p in p + q = – p, we get

    1 + q = – 1

    ⇒ q = – 2

    I HOPE IT HELPS YOU

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