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# Plz Help! I Don't Know How to Do This Question!!!

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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
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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
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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
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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)   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
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Yep..thanks for catching that, guest....correction made !!!   CPhill  Mar 2, 2019