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Question: Solve the quadratic equation \(2{x^2} + 5x + 3 = 0\) using the quadratic formula....

Solve the quadratic equation 2x2+5x+3=02{x^2} + 5x + 3 = 0 using the quadratic formula.

Explanation

Solution

Here we are asked to solve the given quadratic equation that is we have to find its roots. Since it is an equation of order two it will have two roots. The roots of a quadratic equation can be found by using the Quadratic formula.
The word quadratic means second degree values of the given variables.
The formula that we need to know before solving the problem:
Let ax2+bx+c=0a{x^2} + bx + c = 0 be a quadratic equation then the roots of this equation are given by −b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}.

Complete step-by-step solution:
It is given that 2x2+5x+3=02{x^2} + 5x + 3 = 0 we aim to solve this equation, that is we have to find its roots.
We know that the number of roots of an equation is equal to its degree. Here the degree of the given equation is two thus this equation will have two roots.
The roots of a quadratic equation can be found by using the formula −b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} where aa - coefficient of the term x2{x^2}, bb - coefficient of the term xx, and cc - constant term.
First, let us collect the required terms for the formula from the given quadratic equation to solve it.
From the given equation 2x2+5x+3=02{x^2} + 5x + 3 = 0, we have a=2a = 2, b=5b = 5, and c=3c = 3.
On substituting these terms in the formula, we get
x=−(5)±(5)2−4(2)(3)2(2)x = \dfrac{{ - \left( 5 \right) \pm \sqrt {{{\left( 5 \right)}^2} - 4\left( 2 \right)\left( 3 \right)} }}{{2\left( 2 \right)}}
On simplifying this we get
x=−5±25−244x = \dfrac{{ - 5 \pm \sqrt {25 - 24} }}{4}
On further simplification we get
x=−5±14x = \dfrac{{ - 5 \pm \sqrt 1 }}{4}
=−5±14= \dfrac{{ - 5 \pm 1}}{4}
⇒x=−5+14\Rightarrow x = \dfrac{{ - 5 + 1}}{4} and ⇒x=−5−14 \Rightarrow x = \dfrac{{ - 5 - 1}}{4}
On solving the above, we get
⇒x=−32\Rightarrow x = \dfrac{{ - 3}}{2} and ⇒x=−1 \Rightarrow x = - 1
Thus, we got the roots of the given quadratic equation that is ⇒x=−32 \Rightarrow x = \dfrac{{ - 3}}{2} and ⇒x=−1 \Rightarrow x = - 1.

Note: The roots of the equation are nothing but the possible value of the unknown variable in that equation. Also, the number of roots of an equation depends on its degree. The degree of an equation is the highest power of the unknown variable in that equation.
Also, in the quadratic equation, the a=0a = 0 is never possible, because then it will be a linear equation.
Also, note that complex imaginary values ii can be reframed in the real form of i2=−1{i^2} = - 1 or in the inverse form as −1=i\sqrt { - 1} = i.