# Answer to Question #27992 in Differential Equations for Luke

Question #27992

The roots of the equation 4x-squared + kx = 9, differ by 5. Calculate the value(s) of k

Expert's answer

4x^2+kx-9=0. Let's divide both parts of the equation by 4.

x^2+(k/4)x-9/4=0

Let A be one of the roots, then the other is A+5. Using Vieta's formulas, we have:

Now we solve the

If A=-1/2, then k=-16. (form

If A=9/2, then k=16.

x^2+(k/4)x-9/4=0

Let A be one of the roots, then the other is A+5. Using Vieta's formulas, we have:

**(1)**A+A+5=-k/4;**(2)**A*(A+5)=-9/4.Now we solve the

**(2)**equation. A1=-1/2,A2=-9/2. So there are to value of k.If A=-1/2, then k=-16. (form

**(1)**equation)If A=9/2, then k=16.

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