Question

Solve $2{x^2} + 9x - 18 = 0$

Answer 1

Hint: To solve the above given expression, we must know that a quadratic equation has $2$ roots or zeroes. The roots can be found by various methods out of which formula method is easiest.
Formula used:- \[x = \dfrac{{ - b\sqrt {{b^2} - 4ac} }}{{2a}}\]
By using the above given formulae, we can easily find out the roots of a given 2nd degree equation, but we need to keep in mind that this formulae only works when the degree of the equation is 2, which means this formulae cannot be applied to find out the roots of a 3 degree equation.

Complete step by step solution:
The equation we have to solve is $2{x^2} + 9x - 18 = 0$.
Now, By applying the formula method, we get:-
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Where a is the coefficient of ${x^2}$
b is the coefficient of $x$
And c is the coefficient of constant term.
Here,$a = 2,b = 9$ and $c = - 18$
Now,
We first simplify; $D = \sqrt {{b^2} - 4ac} $
Where D is called discriminant.
We get:
$ \Rightarrow \sqrt {{9^2} - \left( {4 \times 2 \times \left( { - 18} \right)} \right)} $
$ \Rightarrow \sqrt {81 + \left( {8 \times 18} \right)} $
$ \Rightarrow \sqrt {81 + 144} $
$ \Rightarrow \sqrt {225} $
Now, as we know that \[225\] is a perfect square of \[15\].
$ \therefore D = 15$, because $D > 0$ {As $D > 0,\,\, \therefore$ we will get two distinct and real roots}
By using the formula method, we get:-
$\Rightarrow x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$
$\Rightarrow \dfrac{{ - 9 + 15}}{{2 \times 2}}$
$ = \dfrac{3}{2}$
OR
$\therefore x = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
$ = \dfrac{{ - 9 - 15}}{{2 \times 2}}$
$ = - 6$
We get $2$ values of $x$ i.e, $x = \dfrac{3}{2}$ or $x = - 6$

Note: In finding solution by formula method, remember to first find value for $\sqrt {{b^2} - 4ac} $, as if $\sqrt {{b^2} - 4ac} = 0$ then we get real and equal roots, if $\sqrt {{b^2} - 4ac} > 0$ then we get 2 real and distinct roots But, If we get $\sqrt {{b^2} - 4ac} < 0$, then the roots will be complex roots.