The equation (t – 1)x^2 + 4x + (t – 5) = 0, where t is a constant has no real roots. Show that t satisfies t2–6t+1>0
This is a quadratic equation of the form ax^2 + bx +c, where (t-1)=a 4=b . (t-5)=0 Therefore if there are no real roots, you get that b^2-4ac<0. (using the quadratic formula)
First plug in values for a b and c: 4^2-4(t-1)(t-5)<0 Secondly rearrange the equation: 16<4(t-1)(t-5)
Then simplify (divide by 4): 4<(t-1)(t-5)
Then expand the brackets: 4<t^2-5t-t+5
Finally rearrange for final result: t^2-6t+1<0
RY
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