Question

If x = 2t²and y =t/2+ 1 then for what value of t, x = 2y holds​

Answer 1

Answer:

$\bf t=\frac{1 \pm\sqrt{17} }{4}$

Step-by-step explanation:

x = 2t²

and y =t/2+ 1

$x=2y\\\\2t^2=2(\frac t2+1)\\\\2t^2=t+2\\\\2t^2-t-2=0\\\\t=\frac{1 \pm\sqrt{(1+4\times4)} }{4} \\\\\bf t=\frac{1 \pm\sqrt{17} }{4}$