Question

Using the quadratic formula, Solve\[2{x^2} - 5x - 3 = 0\].

Answer 1

Hint:Quadratic Formula: $a{x^2} + bx + c = 0$ Here $a,\;b,\;c$are numerical coefficients.

So to solve $x$ we have:$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

So in order to solve the above given question using quadratic formula we have to find the values of
$a,\;b,\;c$corresponding to the given question. Then by substituting the values in the above equation we can find the values for $x$ and thereby solve it.

Complete step by step solution:
Given
\[2{x^2} - 5x - 3 = 0................................\left( i \right)\]

Now we need to compare (i) to the general formula and find the values of unknowns. Then we have to use the equation to find $x$by substituting all the values needed in it and by that way we can solve the equation\[2{x^2} - 5x - 3 = 0\].

So on comparing (i) to the general formula$a{x^2} + bx + c = 0$, we get:
$a = 2,\;b = - 5,\;c = - 3.....................\left( {ii} \right)$

Now to solve for $x$we have $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}......................\left( {iii}
\right)$

Now substituting (ii) in (iii) we get:
\[
\;\;\;\;x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( {2 \times - 3}
\right)} }}{{2 \times 2}} \\
\Rightarrow x = \dfrac{{5 \pm \sqrt {\left( {25} \right) - 4\left( { - 6} \right)} }}{4} \\
\Rightarrow x = \dfrac{{5 \pm \sqrt {\left( {25} \right) - \left( { - 24} \right)} }}{4} \\
\Rightarrow x = \dfrac{{5 \pm \sqrt {\left( {25} \right) + \left( {24} \right)} }}{4} \\
\Rightarrow x = \dfrac{{5 \pm \sqrt {\left( {49} \right)} }}{4} \\
\Rightarrow x = \dfrac{{5 \pm 7}}{4} \\
\]

Now there are two possibilities of$x$, which is produced either by addition or by subtraction. It’s found such that:
\[
\Rightarrow x = \dfrac{{5 + 7}}{4}\;\;\;\;\;{\text{and}}\;\;\;x = \dfrac{{5 - 7}}{4} \\

\Rightarrow x = \dfrac{{12}}{4}\;\;\;\;\;\;\;\;{\text{and}}\;\;\;x = \dfrac{{ - 2}}{4} \\
\Rightarrow x = 3\;\;\;\;\;\;\;\;\;\;{\text{and}}\;\;\;x = - 0.5 \\
\\
\Rightarrow x = 3,\; - 0.5 \\
\]

Therefore on solving \[2{x^2} - 5x - 3 = 0\]we get\[x = 3,\; - 0.5\].

Additional Information: In order to check if the values of $x$that are obtained are correct or not we simply have to substitute the values of $x$in the given parent equation and see whether the equation is satisfied or not.

So here \[x = 3,\; - 0.5\]

Substituting each values of \[x\] in the equation (i) which is: \[2{x^2} - 5x - 3 = 0\]

So let\[x = 3\]:
\[
\Rightarrow 2{x^2} - 5x - 3 = 2{\left( 3 \right)^2} - \left( {5 \times 3} \right) - 3 \\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \left( {2 \times 9} \right) - 15 - 3 \\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 18 - 18 \\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.....................\left( {iv} \right) \\
\]

So (iv) implies that the equation is mathematically correct, such that \[x = 3\]is a correct value.

Now let\[x = - 0.5\]:
\[
\Rightarrow 2{x^2} - 5x - 3 = 2{\left( { - 0.5} \right)^2} - \left( {5 \times \left( { - 0.5} \right)} \right) -
3 \\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \left( {2 \times 0.25} \right) - \left( { - 2.5} \right) - 3 \\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.5 + 2.5 - 3 \\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 3 - 3 \\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.....................\left( v \right) \\
\]

So (v) implies that the equation is mathematically correct, such that \[x = - 0.5\]is a correct value.

Note:Quadratic formula is mainly used in conditions where grouping method cannot be used or when the polynomial cannot be reduced into some general identity. Quadratic formula method is an easier and direct method in comparison to other methods.

Also while using the Quadratic formula when $\sqrt {{b^2} - 4ac} $is a negative root then the corresponding answer would be a complex number.