Question

for what value of k are the roots of 2x^2-8x+k=0 equal?

Answer 1

2x^2-8x+k=0
Solve the above for x.\[\left\{x=\frac{1}{2} \left(4-\sqrt{2} \sqrt{8-k}\right)\right\},\left\{x=\frac{1}{2} \left(4+\sqrt{2} \sqrt{8-k}\right)\right\} \]\[\frac{1}{2} \left(4-\sqrt{2} \sqrt{8-k}\right)\text{=}\frac{1}{2} \left(4+\sqrt{2} \sqrt{8-k}\right) \]\[\frac{1}{2} \left(4-\sqrt{2} \sqrt{8-k}\right)-\left(\frac{1}{2} \left(4+\sqrt{2} \sqrt{8-k}\right)\right)\text{= }0\]\[-\sqrt{2} \sqrt{8-k}=0 \]\[\left(-\sqrt{2} \sqrt{8-k}\right)^2=0^2\]\[k = 8 \]\[2 x^2- 8 x+8 \text{=} 0\]\[2 (x-2)^2=0 \]