Question

The point $$(t^2+2t+5,2t^2+t-2)$$ lies on the line $$x+y=2$$ for
A. All real values of $$t$$
B. Some real values of $$t$$
C. $$t=(-3\pm \sqrt{3})/6$$
D. No real values of $$t$$

Answer 1

Hint: To solve this problem, first we need to understand the Sridharacharya method and then we have to substitute the values of given point in the given equation and then solve the quadratic equation using Sridharacharya method and then compare the Discriminant and you will get your required answer.
Complete step-by-step solution:
Before solving the given problem, first let’s understand the Sridharacharya method:
 Sridhara wrote down rules for solving quadratic equations, it is the most common method of finding the roots of the quadratic equation and it is known as Sridharacharya rule.
Let’s consider a quadratic equation: $$a{x}^2+bx+c$$ , where $$a,b,c$$ are real numbers and numeral coefficients. And here $$a\ne 0$$ , because if it is equal to zero then the equation will not remain quadratic anymore and it will become a linear equation.
This formula for a quadratic equation is used to find the roots of the equation. Since, quadratic equations have a degree equal to two, hence there will be two solutions for the equation.
The formula to find the roots of this equation will be:
$$x={\displaystyle \frac{[-b\pm \sqrt{(b^2-4ac)}]}{2a}}$$
In this, the nature of the roots depends on $$(b^2-4ac)$$ . This is called as the Discriminant and it is represented as $$D$$
Therefore, Discriminant $$D=b^2-4ac$$
There are three cases after finding the value of $$D$$
Case1:- When $$D>0$$ , then the roots are real and distinct in nature.
 Case2:- When $$D=0$$ , then the roots are real and equal in nature.
Case3:- When $$D<0$$ , then the roots are not real.
As we are given in the question:
$$x+y=2$$ $$..........(1)$$
As we are given that the points $$(t^2+2t+5,2t^2+t-2)$$ lies on the line
So, we will substitute the values of given point in the equation $$(1)$$
$$\begin{array}{rl}& x+y=2\\ & \Rightarrow t^2+2t+5+2t^2+t-2=2\end{array}$$
$$\Rightarrow 3t^2+3t+1=2$$
Now we will use Sridharacharya method to solve this equation:
So, $$D=b^2-4ac$$
$$\Rightarrow D={(3)}^2-4\times 3\times 1$$
$$\Rightarrow D=9-12$$
$$\Rightarrow D=-3$$
A, we can see that $$D<0$$
$$\Rightarrow -3<0$$
So, from above conditions we can say that no real values of $$t$$ exists
Hence, the correct option is $$D$$.

Note: A quadratic equation makes a parabola when graphed on a coordinate plane. If the value of Discriminant is positive, then the graph crosses the x-axis, When the value of Discriminant is zero, then the graph touches the x-axis but does not cross it, When the value of Discriminant is negative, then the graph does not even touch the x-axis.