Question

What is the condition so that the point $\left( {{t^2} + 2t + 5, 2{t^2} + t - 2} \right)$ lies on the line $x + y = 2$?
A. All real values of $t$
B. Some real values of $t$
C. $t = \dfrac{{\left( { - 3 \pm \sqrt 3 } \right)}}{6}$
D. No real values of $t$

Answer 1

Hint: First, substitute the coordinates of the point in the equation of a line. Then simplify the equation. The new equation of a line is a quadratic equation of variable $t$. Use the quadratic formula and get the values of $t$.

Formula Used:
The quadratic formula: The roots of a quadratic equation $a{x^2} + bx + c = 0, a \ne 0$ are $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
The given equation of a line is $x + y = 2$ and the point is $\left( {{t^2} + 2t + 5, 2{t^2} + t - 2} \right)$.
If the point lies on the line, then the point satisfies the equation of the line.
Substitute the coordinates of the point in the equation of the line.
We get,
$\left( {{t^2} + 2t + 5} \right) + \left( {2{t^2} + t - 2} \right) = 2$
$ \Rightarrow 3{t^2} + 3t + 3 = 2$
$ \Rightarrow 3{t^2} + 3t + 1 = 0$
This is a quadratic equation with a variable $t$.

Now use the quadratic formula to find the value of the variable $t$.
$t = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4\left( 3 \right)\left( 1 \right)} }}{{2\left( 3 \right)}}$
$ \Rightarrow t = \dfrac{{ - 3 \pm \sqrt {9 - 12} }}{6}$
$ \Rightarrow t = \dfrac{{ - 3 \pm \sqrt { - 3} }}{6}$
$ \Rightarrow t = \dfrac{{ - 3 \pm i\sqrt 3 }}{6}$ [ Since $\sqrt { - n} = i\sqrt n $ ]
Thus, the roots of the equation are complex numbers.
Means, the variable $t$ has no real values.

Option ‘D’ is correct

Note: The roots of the quadratic equation are:
The discriminant of the quadratic formula is $D = {b^2} - 4ac$.
If $D = 0$, then the roots are real and equal.
If $D > 0$, then the roots are real and unequal.
If $D < 0$, then the roots are not real.