SOLUTION: Please help me on this problem 4X+4X+1=0 how many real solutions dose this problem have

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Question 972451: Please help me on this problem 4X+4X+1=0 how many real solutions dose this problem have
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
if you have:
4x%2B4x%2B1=0+, then
8x%2B1=0+
8x=-1+
x=-1%2F8+=> one real solution
+graph%28600%2C+600%2C+-5%2C+5%2C+-5%2C+5%2C+8x%2B1%29+


but, if you have
4x%5E2%2B4x%2B1=0+, then
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-4+%2B-+sqrt%284%5E2-4%2A4%2A1+%29%29%2F%282%2A4%29+
x+=+%28-4+%2B-+sqrt%2816-16+%29%29%2F8+
x+=+%28-4+%2B-+sqrt%280+%29%29%2F8+
x+=+%28-4%2F8+%2B-+0%2F8%29+
x+=+%28-cross%284%291%2Fcross%288%292+%2B-+0%29+
x+=+%28-1%2F2+%2B-+0%29+
x+=+-1%2F2++=>one real solution
+graph%28600%2C+600%2C+-5%2C+5%2C+-5%2C+5%2C+4x%5E2%2B4x%2B1%29+