Question

How do you solve \[2{x^2} + 5x = 0\] using the quadratic formula?

Answer 1

Hint: Use method of determinant to solve for the value of x from the given quadratic equation. Compare the quadratic equation with general quadratic equation and substitute values in the formula of finding roots of the equation.
* For a general quadratic equation \[a{x^2} + bx + c = 0\], roots are given by formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step-by-step answer:
We are given the quadratic equation \[2{x^2} + 5x = 0\] … (1)
We know that general quadratic equation is \[a{x^2} + bx + c = 0\]
On comparing with general quadratic equation \[a{x^2} + bx + c = 0\], we get \[a = 2,b = 5,c = 0\]
Substitute the values of a, b and c in the formula of finding roots of the equation.
\[ \Rightarrow x = \dfrac{{ - (5) \pm \sqrt {{{(5)}^2} - 4 \times 2 \times 0} }}{{2 \times 2}}\]
Square the values inside the square root in numerator of the fraction
\[ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {{5^2}} }}{{2 \times 2}}\]
Cancel square root by square power in the numerator
\[ \Rightarrow x = \dfrac{{ - 5 \pm 5}}{{2 \times 2}}\]
So, \[x = \dfrac{{ - 5 + 5}}{{2 \times 2}}\]and \[x = \dfrac{{ - 5 - 5}}{{2 \times 2}}\]
I.e. \[x = \dfrac{0}{{2 \times 2}}\]and \[x = \dfrac{{ - 10}}{{2 \times 2}}\]
Cancel possible factors from numerator and denominator
\[x = 0\]and \[x = \dfrac{{ - 5}}{2}\]
i.e. \[x = 0\]and \[x = - 2.5\]

\[\therefore \]Solution of the equation \[2{x^2} + 5x = 0\] is \[x = 0\]and \[x = - 2.5\].

Note:
Many students leave their answer in fraction form, which is okay as the answer can be according to the requirement of the question, but keep in mind we should always solve the simple fractions like these to write the final answer in decimal terms. Also, when writing the roots students don’t write the value 0 as root which is wrong, since x comes out to be 0, it will be the root of the quadratic equation.
Alternate method:
We have to simplify the equation \[2{x^2} + 5x = 0\]
Take x common from both terms
\[ \Rightarrow x(2x + 5) = 0\]
We know products with two values can be zero when one of them is 0 or both are zero. We equate both values to 0.
\[ \Rightarrow x = 0\]and \[2x + 5 = 0\]
Shift constant to RHS
\[ \Rightarrow x = 0\]and \[2x = - 5\]
Cross multiply the value from LHS to denominator of RHS
\[ \Rightarrow x = 0\]and \[x = \dfrac{{ - 5}}{2}\]
\[ \Rightarrow x = 0\]and \[x = - 2.5\]
\[\therefore \]Solution of the equation \[2{x^2} + 5x = 0\] is \[x = 0\]and \[x = - 2.5\].