91Ó°ÊÓ

We have given the following system of equations :-

$\left\{\begin{array}{l}2y^2-x=0\\ y=x+1\end{array}\right.$

We have to solve this system of equations. That is we have to find pair for values of $xandy$that satisfies this system of equations.

We will substitute the value of$y$from second equation in first one. Than we have a quadratic equation in$x$. After that we will solve the equations, to get required values.

The given system of equations is :-

$\left\{\begin{array}{l}2y^2-x=0\\ y=x+1\end{array}\right.$

Substitute value of $y=x+1$in the first equation $2y^2-x=0$, then we have :-

$$\begin{gathered} 2{\left(x+1\right)}^2-x=0 \\ 2(x^2+2x+1)-x=0 \\ 2x^2+4x+2-x=0 \\ 2x^2+3x+2=0 \end{gathered}$$

We have :-

$2x^2+3x+2=0$.

Here this does not factorized easily. So we check the discriminant.

The discriminant of a quadratic equation is $D=b^2-4ac$, where $b$is the middle term and $aandc$are first and last terms respectively.

Then the discriminant of above quadratic equation is :-

$$\begin{gathered} D={\left(3\right)}^2-(4×2×2) \\ ⇒D=9-16 \\ ⇒D=-7 \end{gathered}$$

Here D is negative. That is there is no real solution of this quadratic equation.

So we can conclude that given system has no solution.