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For what values of \(a\) does the equation \((a^{2}+2a)x^{2}+(3a)x+1=0\) yield no real solutions \(x\)? Express your answer in interval notation.

 Mar 1, 2019
 #1
avatar+99324 
+2

What have you tried?

I'd try using the quadratic formula. The bit under the square must be great or equal to zero or else the answer/s will not be real.

 Mar 1, 2019
 #2
avatar+777 
+1

try subbing it in formula b^2-4ac < 0 ( this represents when a quadratic formula has no real solution)

then rearrange it 

 Mar 1, 2019
 #3
avatar+98130 
+1

We will have no real solutions if the discriminant is < 0

 

So

 

(3a)^2 - 4(a^2 + 2a)(1) <  0

 

9a^2 - 4a^2 - 8a < 0

 

5a^2 -8a < 0       factor

 

a ( 5a -8) < 0

 

This will =  0       whenever a = 0    or   a = 8/5

 

If a > 8 / 5....it will be positive

And if a < 0  it will be positve

 

So.....the values of a that produce no real solutions are   when a is on the interval (0, 8/5)

 

 

cool cool cool

 Mar 2, 2019
edited by CPhill  Mar 2, 2019
 #4
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Wait quick question, shouldn't it be \(5a^2-8a<0\) because it's \(b^2-4ac<0\) which comes out to \((3a)^2-4(a^2+2a)\) which is \(9a^2-4a^2-8a=5a^2-8a\) instead of \(5a^2-4a\) ?

Guest Mar 2, 2019
 #5
avatar+98130 
0

Yep..thanks for catching that, guest....correction made !!!

 

cool cool cool

CPhill  Mar 2, 2019

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