Answered

Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

If y = x² + kr - k, for what values of k will the quadratic have two real solutions?

Sagot :

The given expression is

[tex]y=x^2+kr-r[/tex]

Where a = 1, b = 0, and c = kr-k.

Let's use the discriminant

[tex]\begin{gathered} D=b^2-4ac \\ D=0^2-4\cdot1\cdot(kr-r) \\ D=-4kr+4r \end{gathered}[/tex]

It is important to know that the equation has two real solutions when the discriminant is greater than zero, so

[tex]-4kr+4r>0[/tex]

Let's factor out the greatest common factor

[tex]4r(-k+1)>0[/tex]

Now, we solve for k.

[tex]\begin{gathered} -k+1>\frac{0}{4r} \\ -k+1>0 \\ -k>-1 \\ k<1 \end{gathered}[/tex]

Hence, k must be less than 1 in order to have two real solutions.

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.