Question

If x=-2 and x=-1/5 are solution of the equation 5x^2+px+r=0 find the value of p and r

Answer 1

Heya mate,Here is ur answer

x=-2

P(x) = 5x^2 +px + r

P(-2) = 5 × (-2)^2 + p×-2 + r

0= 5×4 -2p +r

0= 20-2p +r

2p-r = 20-----(1)

x=-1/5

P(-1/5)= 5×(-1/5)^2 +p×(-1/5) +r

$0 = 5 \times \frac{1}{25} - \frac{p}{5} + r$


$0 = \frac{1}{5} - \frac{p}{5} + r$


$0 = \frac{1 - p + 5r}{5}$


$0 \times 5 = 1 - p + 5r$


$0 = 1 - p + 5r$

$p - 5r = 1$

Multiplying by 2

$2(p - 5r) = 5 \times 1$


$2p - 10r = 5$

So, the two equations are

2p-r=20
2p-10r=5
- + -
-------------
9r =15
--------------

$r = \frac{15}{9}$


r=5/3

2p-r =20

$2p - \frac{5}{3} = 20$


$\frac{6p - 5}{3} = 20$


$6p - 5 = 20 \times 3$



$6p - 5 = 60$


6p=60+5

6p=65

p=65/6

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@Laughterqueen

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Answer 2

Answer:

p = 65/6

Step-by-step explanation: