Question

solve the differential equation p^2+7p+12=0

Answer 1

it's easy. there needs to be split the middle term.

we have
p^2+7p+12=0

first, divide the constant term with the the first term. doing this we get 12p^2

find two such numbers whose product is +12p^ and sum is +7

+3p and +4p r the numbers
[4p+3p=7p, 4p*3p=12p^2]

putting 4p and 3p in the place of 7p, we get

p^2+4p+3p+12
p[p+4]+3[p+4][taking common from both terms]
[p+3][p+4]

HOPE THIS ANSWER HELPES!!!!

Answer 2

Answer:

The solution is p= -4, -3.

Step-by-step explanation:

Given, $p^{2}+7p+12=0$

To solve the equation.

By middle term splitting method we have:

$p^{2}+7p+12=0\\p^{2}+4p+3p+12=0\\p(p+4)+3(p+4)=0\\(p+4)(p+3)=0$

By zero property of product we have:

p+4=0 or p+3=0

p= -4, -3

Hence, the solution of $p^{2}+7p+12=0$ is p= -4, -3.

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