## 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

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## Answers ( )

Step-by-step explanation:Given –

To Find –

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.

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