SOLUTION: Under square root(3-x) + under square root(x+5)=4

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Question 289118: Under square root(3-x) + under square root(x+5)=4
Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
Under square root(3-x) + under square root(x+5)=4
I think you mean the following:
+sqrt%283-x%29+%2B+sqrt%28x%2B5%29+=+4+
set a = 3 - x
set b = x + 5
+sqrt%28a%29+%2B+sqrt%28b%29+=+4+ (square both sides)
+a+%2B+2sqrt%28a%29sqrt%28b%29+%2B+b+=+16+
+a+%2B+2sqrt%28ab%29+%2B+b+=+16+
+2sqrt%28ab%29+=+16+-+a+-+b+ (square both sides)
+4ab+=+%2816+-+a+-+b%29%5E2+ (evaluate (16 - a - b)^2)
+%2816+-+a+-+b%29%2816+-+a+-+b%29+
+256+-+16a+-+16b+-+16a+%2B+a%5E2+%2B+ab+-+16b+%2B+ab+%2B+b%5E2+
+a%5E2+%2B+b%5E2+%2B+2ab+-+32a+-+32b+%2B+256+ (returning)
+4ab+=+%2816+-+a+-+b%29%5E2+
+4ab+=+a%5E2+%2B+b%5E2+%2B+2ab+-+32a+-+32b+%2B+256+
+0+=+a%5E2+%2B+b%5E2+-+2ab+-+32a+-+32b+%2B+256+
+a%5E2+=+%283+-+x%29%5E2+
+a%5E2+=+9+-+6x+%2B+x%5E2+
+a%5E2+=+x%5E2+-+6x+%2B+9+
+b%5E2+=+%28x+%2B+5%29%5E2+
+b%5E2+=+x%5E2+%2B+10x+%2B+25+
+ab+=+%283+-+x%29%28x+%2B+5%29+
+ab+=+3x+%2B+15+-+x%5E2+-+5x+
+ab+=+-x%5E2+-+2x+%2B+15+


+0+=+2x%5E2+%2B+4x+%2B+34+%2B+%282x%5E2+%2B+4x+-+30%29+-+256+%2B+256+
+0+=+4x%5E2+%2B+8x+%2B+4+
+0+=+x%5E2+%2B+2x+%2B+1
+0+=+%28x+%2B+1%29%28x+%2B+1%29+
x = -1
check: +sqrt%283-x%29+%2B+sqrt%28x%2B5%29+=+4+
+sqrt%283+-+-1%29+%2B+sqrt%28-1+%2B5%29+
+sqrt%284%29+%2B+sqrt%284%29+
+2+%2B+2+=+4+