SOLUTION: For what real values of a does the quadratic 4x^2 + ax + 25 have non-real roots? Give your answer as an interval.

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Question 1075729: For what real values of a does the quadratic 4x^2 + ax + 25 have non-real roots? Give your answer as an interval.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The quadratic formula says that the roots are:
+x+=+%28+-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
when the equation has the form:
+y+=+a%2Ax%5E2+%2B+b%2Ax+%2B+c+
They decided that +b+=+a+ ( a little confusing )
so the list is:
+a+=+4+
+b+=+a+
+c+=+25+
Now I can say:
+x+=+%28+-a+%2B-+sqrt%28+a%5E2-4%2A4%2A25+%29%29%2F%282%2A4%29+
+x+=+%28+-a+%2B-+sqrt%28+a%5E2+-+400+%29%29%2F8+
The roots will be non-real ( imaginary ) whenever
+a%5E2+-+400+ is negative. ( this is called the discriminant )
So I can say:
+a%5E2+-+400+%3C+0+
+a%5E2+%3C+400+
+a+%3C+20+
and also:
+a+%3E+-20+
Putting these together:
+-20+%3C+a+%3C+20++
------------------------
example:
+a+=+0+ satisfies the conditions, so
+x+=+%28+-a+%2B-+sqrt%28+a%5E2+-+400+%29%29%2F8+
+x+=+%28+-0+%2B-+sqrt%28+0%5E2+-+400+%29%29%2F8+
+x+=+%281%2F8%29%2A%28+sqrt%28+-400+%29+%29+
+x+=+%281%2F8%29%2A%28+20i+%29+
+x+=+%285%2F2%29%2Ai+
and also:
+x+=+%281%2F8%29%2A%28+-sqrt%28+-400+%29+%29+
+x+=+%28-5%2F2%29%2Ai+
Here is the plot when +a+=+0+
+graph%28+400%2C+400%2C+-6%2C+6%2C+-5%2C+50%2C+4x%5E2+%2B+25+%29+