Question

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Answer 1

5x^2+px+q=0

Or, 5*-2^2+p*-2+q=0....... 5*-1/5^2+-p/5+q=0

Or, 20+-2p+q=0............... 1/5+-p/5+q=0

Or, -2p+q= -20.................... -p/5+q=-1/5

Or,p+q=20/-2= 10...............Or, p+q=1

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Answer 2

Answer:

p = 11 and q = 2

Step-by-step explanation:

We have,

5x² + px + q = 0

also,

x = (-2) and x = (-1/5)

Let p(x) = 5x² + px + q

Now, when x = (-2)

p(-2) = 5(-2)² + p(-2) + q

As x = (-2) is a solution, p(-2) = 0

0 = 5(4) - 2p + q

0 = 20 - 2p + q

2p - 20 = q

q = 2p - 20 ----- 1

Also,

p(-1/5) = 5(-1/5)² + p(-1/5) + q

0 = 5(1/25) - (p/5) + q

0 = (1/5) - (p/5) + q

0 = (1/5) - (p/5) + (5q/5)

Now multiplying the whole equation by 5,

1 - p + 5q = 0

p - 5q = 1 ------ 2

Substituting eq.1 in eq.2,

p - 5(2p - 20) = 1

p - 10p + 100 = 1

-9p = 1 - 100

-9p = -99

p = (-99/-9)

p = 11

Putting p = 11, in eq.1,

q = 2(11) - 20

q = 22 - 20

q = 2

Hence,

p = 11 and q = 2

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