How do you solve #2p ^ { 2} + 49= - 21p#?

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Answer 1 · 2017-12-08T11:47:49

See a solution process below:

First, add #color(red)(21p)# to each side of the equation to put the equation in standard form:

#2p^2 + color(red)(21p) + 49 = - 21p + color(red)(21p)#

#2p^2 + 21p + 49 = 0#

Next, factor the left side of the equation as:

#(2p + 7)(p + 7) = 0#

Now, solve each term on the left side of the equation for #0# to find the solutions to the problem:

Solution 1:

#2p + 7 = 0#

#2p + 7 - color(red)(7) = 0 - color(red)(7)#

#2p + 0 = -7#

#2p = -7#

#(2p)/color(red)(2) = -7/color(red)(2)#

#(color(red)(cancel(color(black)(2)))p)/cancel(color(red)(2)) = -7/2#

#p = -7/2#

Solution 2:

#p + 7 = 0#

#p + 7 - color(red)(7) = 0 - color(red)(7)#

#p + 0 = -7#

#p = -7#

The Solutions Are: #p = {-7/2, -7}#