Question

Find the values of x which satisfies the equation ${{x}^{2}}+3x-18$?

Answer 1

Hint: The given equation is a quadratic equation so it will have at the most $2$ solution which can be real or imaginary depending on the discriminant value of the given equation.

Complete step by step answer:
Quadratic equations are those which have the maximum/highest degree or power of the variable is $2$.
For solving Quadratic equation we first should check its determinant(D) value which for a quadratic equation $a{{x}^{2}}+bx+c$is given by the equation $D={{b}^{2}}-4ac$, where b = coefficient of x , a = coefficient of b and c = a constant
Case 1 – if D >$0$,than x has $2$distinct real solution;
Case 2 – if D=$0$,than x has $2$equal real solution ;
Case 3 – if D<\[0\],than x has $2$ imaginary solution that will be conjugate of each other for ex p+iq and p – iq;
For the equation${{x}^{2}}+3x-18$ ,a = $1$,b=$3$,c =-$18$, now putting these values in the formula for D we get
 $\begin{align}
  & D={{3}^{2}}-4*1*(-18) \\
 & D=81 \\
\end{align}$
So as D value is >$0$ we can say the given solution has 2 distinct real solutions now we can write the equation as -
$\begin{align}
  & {{x}^{2}}+6x-3x-18=0 \\
 & x(x+6)-3(x+6)=0 \\
 & (x+6)(x-3)=0 \\
 & x+6=0 \\
 & x=-6 \\
 & or \\
 & x-3=0 \\
 & x=3; \\
\end{align}$
So we calculated the value of x can be $3,-6$ which satisfies the equation ,we used $3,6$
As they are factor of $18$ so that we can take common values you can take any factor of $18$
But they should be such that we can take common values like we did for this equation .

Note: The maximum number of real distinct solutions of a polynomial depends on the highest degree of unknown variable .