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Please help, need both 3 and 1, show your work if possible

Please help, need both 3 and 1, show your work if possible-example-1

1 Answer

2 votes

Answer:

1. The equation has 2 solutions 3. p = 0, p = -2

Explanation:

1. The quadratic equation is ax^2+bx+c=0
x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)

in the equation a = 2, b = 3 and c = 5

so the equation would be
x_(1,\:2)=(-(-3)+ √((-3)^2-4*2*5))/(2*2)


x_(1,\:2)=(3+√(9-4*2*5))/(2*2)

simplify


3+√(31)i

divide by 4


(3)/(4) +(√(31) )/(4)i or
x=(-\left(-3\right)-√(\left(-3\right)^2-4\cdot \:2\cdot \:5))/(2\cdot \:2):\quad (3)/(4)-i(√(31))/(4)

3. 4p(5p + 10) = 0 Using the Zero Factor Principle aka. zero-product property If a*b=0 then a=0 or b=0 or both a=0 and b=0

p = 0

5p + 10 = 0

subtract ten from both sides

5p = -10

divide by 5

p = -2

User Adam Balsam
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