SOLUTION: if the number p is 7 more than the number q and the sum of the squares of p and q is 85 find the product of p and q?
Algebra.Com
Question 983423: if the number p is 7 more than the number q and the sum of the squares of p and q is 85 find the product of p and q?
Found 2 solutions by rothauserc, Cromlix:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
p = q + 7
(q + 7)^2 + q^2 = 85
q^2 + 14q + 49 + q^2 = 85
2q^2 + 14q + 49 = 85
2q^2 + 14q - 36 = 0
q^2 + 7q - 18 = 0
(q+9) * (q-2) = 0
q is -9 or 2, therefore
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we have two solutions
q = -9 and p = -2
q = 2 and p = 9
Answer by Cromlix(4381) (Show Source): You can put this solution on YOUR website!
Hi there,
If p is 7 more than q
p = q + 7
the sum of the squares of p and q is 85
p^2 + q^2 = 85
Substitute q + 7 for 'p'
(q + 7)^2 + q^2 = 85
Multiply out
q^2 + 14q + 49 + q^2 = 85
Collect like terms
q^2 + q^2 + 14q + 49 - 85 = 0
2q^2 + 14q - 36 = 0
Divide thro'out by 2
q^2 + 7q - 18 = 0
Factorise
(q - 2)(q + 9) = 0
So, either;
q - 2 = 0
q = 2
OR
q + 9 = 0
q = -9
........
Using q = 2
p = q + 7
p = 9
pq = 9 x 2
pq = 18
........
Using q = -9
p = q + 7
p = -2
pq = 9 x -2
pq = -18
.........
Hope this helps:-)
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