Question

The point $\left(t^2+2t+5,2t^2+t-2\right)$ lies on the line $x+y=2$ for

• A. All real values of $t$
• B. Some real values of $t$
• C. $t=\frac{\left(-3\pm \sqrt{3}\right)}{6}$
• D. No real values of $t$

Answer 1

The correct option is D

No real values of $t$

Explanation for the correct answer:

Solve the given equation of line based on given information

In the question, it is given that point $\left(t^2+2t+5,2t^2+t-2\right)$ lies on the line $x+y=2$.

So, the point satisfies the given line, such that, $x=t^2+2t+5,y=2t^2+t-2$,

$$\begin{gathered} t^2+2t+5+2t^2+t-2=2 \\ \Rightarrow 3t^2+3t+3=2 \\ \Rightarrow 3t^2+3t+1=0 \end{gathered}$$

We know that the discriminant of the above quadratic equation is given by, $D=b^2-4ac$

So,

$$\begin{gathered} D=3^2-4\left(3\right)\left(1\right) \\ \Rightarrow D=9-12<0 \end{gathered}$$

Therefore, there is no real value of $t$ for which the given point will lie on the given line.

Hence, option $D$ is the correct answer.