SOLUTION: Please help me solve this equation: {{{sqrt(4p+5) + sqrt (p+5) = 3}}}

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Question 56523: Please help me solve this equation: sqrt%284p%2B5%29+%2B+sqrt+%28p%2B5%29+=+3
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%284p%2B5%29+%2B+sqrt+%28p%2B5%29+=+3
sqrt%284p%2B5%29%2Bsqrt%28p%2B5%29-sqrt%28p%2B5%29=3-sqrt%28p%2B5%29
sqrt%284p%2B5%29=3-sqrt%28p%2B5%29
%28sqrt%284p%2B5%29%29%5E2=%283-sqrt%28p%2B5%29%29%5E2 The rule is highlight%28%28sqrt%28a%29%29%5E2=a%29
4p%2B5=%283-sqrt%28p%2B5%29%29%283-sqrt%28p%2B5%29%29 FOIL

4p%2B5=9-3sqrt%28p%2B5%29-3sqrt%28p%2B5%29%2B%28p%2B5%29
4p%2B5=-6sqrt%28p%2B5%29%2Bp%2B14
4p-p%2B5-14=-6sqrt%28p%2B5%29%2Bp-p%2B14-14
3p-9=-6sqrt%28p%2B5%29
3%28p-3%29=-6sqrt%28p%2B5%29
%283%2F3%29%28p-3%29=%28-6%2F3%29sqrt%28p%2B5%29
%28p-3%29=-2%2Asqrt%28p%2B5%29
%28%28p-3%29%29%5E2=%28-2%2Asqrt%28p%2B5%29%29%5E2
p%5E2-6p%2B9=4%28p%2B5%29 This is a quadratic equation and needs to be set =0 and factored if possible.
p%5E2-6p%2B9=4p%2B20
p%5E2-6p-4p%2B9-20=4p-4p%2B20-20
p%5E2-10p-11=0
(p________)(p______)=0
We need to fill the blanks with two integers that multiply to get -11, but add to get -10:
(-11)*(+1)=-11
(-11)+(+1)=-10
(p-11)(p+1)=0 Set each parenthesis =0 and solve for x:
p-11=0
p=11
P+1=0
p=-1
So x=11 and x=-1 are possible solutions. However, with even radical equations , rational equations, and log equations we have to check our answers for false solutions called extraneous solutions.
substituting x=11
sqrt%284%2Ahighlight%2811%29%2B5%29%2Bsqrt%28highlight%2811%29%2B5%29=3
sqrt%2844%2B5%29%2Bsqrt%2811%2B5%29=3
sqrt%2849%29%2Bsqrt%2816%29=3
7%2B4=3
11=3 This proves that x=11 is an extraneous solution.
Substituting x=-1
sqrt%284%2Ahighlight%28-1%29%2B5%29%2Bsqrt%28highlight%28-1%29%2B5%29=3
sqrt%28-4%2B5%29%2Bsqrt%28-1%2B5%29=3
sqrt%281%29%2Bsqrt%284%29=3
1%2B2=3
3=3 This proves that x=-1 is a solution to the equation:
highlight%28x=-1%29
Happy Calculating!!!