We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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.

Guest Mar 1, 2019

#1**+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.

Melody Mar 1, 2019

#2**+1 **

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

then rearrange it

YEEEEEET Mar 1, 2019

#3**+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)

CPhill Mar 2, 2019