How To Derive Quadratic Formula

An piece of cake way to solve quadratic equations is through factoring. All the same, many of the quadratic equations cannot be factored easily. In these cases, we can use the general quadratic formula to solve whatsoever quadratic equation.

In this commodity, we will larn how to derive the quadratic formula step past pace. Then, we will use information technology to solve quadratic equations.

ALGEBRA

Steps to quadratic formula and exercises

Relevant for…

Learning to derive the quadratic formula stride by step.

Come across steps

ALGEBRA

Relevant for…

Learning to derive the quadratic formula stride by footstep.

See steps

How to derive the quadratic formula?

The quadratic formula $latex x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$ is derived using steps like to those we use to complete the square. This formula takes into account the fact that any equation of the form $latex a{{10}^ii}+bx+c=0$ can be solved to find its roots.

The roots of the equation are points where the graph of the equation crosses the ten-axis.

Therefore, to find the roots of the equation $latex a{{ten}^2}+bx+c=0$, we are going to employ the steps used in completing the square to obtain the general formula:

Step 1: Write the equation in the general class $latex a{{10}^ii}+bx+c=0$, wherea, b,and c are existent numbers, anda≠0.

Pace 2: Move the constantc to the right paw side of the equation:

$latex a{{10}^2}+bx+c-c=0-c$

$latex a{{ten}^2}+bx=-c$

Footstep three: Carve up the equation past the coefficienta:

$latex \frac{one}{a}\left( {a{{10}^{2}}+bx=-c} \right)$

$latex {{10}^{2}}+\frac{b}{a}10=-\frac{c}{a}$

Step 4: Identify the coefficient $latex \frac{b}{a}$, divide it by 2, square it and simplify it:

$latex {{\left( {\frac{{\frac{b}{a}}}{ii}} \right)}^{2}}={{\left( {\frac{b}{{2a}}} \right)}^{2}}=\frac{{{{b}^{two}}}}{{4{{a}^{2}}}}$

Step 5: Add the result of pace iv to both sides of the equation:

$latex {{ten}^{2}}+\frac{b}{a}x+\frac{{{{b}^{2}}}}{{4{{a}^{ii}}}}=-\frac{c}{a}+\frac{b}{{iv{{a}^{2}}}}$

Step 6: Simplify the right hand side of the equation:

$latex {{ten}^{ii}}+\frac{b}{a}10+\frac{{{{b}^{2}}}}{{4{{a}^{two}}}}=-\frac{c}{a}\left( {\frac{{4a}}{{4a}}} \correct)+\frac{{{{b}^{2}}}}{{4{{a}^{ii}}}}$

$latex =-\frac{{4ac}}{{four{{a}^{2}}}}+\frac{{{{b}^{2}}}}{{four{{a}^{2}}}}$

$latex {{x}^{2}}+\frac{b}{a}x+\frac{{{{b}^{2}}}}{{iv{{a}^{2}}}}=\frac{{{{b}^{2}}-4ac}}{{iv{{a}^{ii}}}}$

Step 7: Form the square of the binomial in the expression on the left:

$latex {{\left( {x+\frac{b}{{2a}}} \right)}^{2}}=\frac{{{{b}^{2}}-4ac}}{{4{{a}^{2}}}}$

Step 8: Have the square root of both sides and simplify without forgetting the ± sign on the right side:

$latex x+\frac{b}{{2a}}=\pm \sqrt{{\frac{{{{b}^{2}}-4ac}}{{iv{{a}^{2}}}}}}$

Step nine: Isolate thex on the left and simplify:

$latex x+\frac{b}{{2a}}-\frac{b}{{2a}}=\pm \frac{{\sqrt{{{{b}^{2}}-4ac}}}}{{2a}}-\frac{b}{{2a}}$

$latex 10=-\frac{b}{{2a}}\pm \frac{{\sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$

$latex x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$

How to solve quadratic equations with the quadratic formula?

To solve quadratic equations with the full general quadratic formula, nosotros accept to commencement past organizing the equation in the form $latex a{{ten}^2}+bx+c=0$. Once nosotros have the equation written that way, we just plug in the coefficientsa, b, andc into the quadratic formula:

$latex x=\frac{{-b\pm \sqrt{{{{b}^{ii}}-4ac}}}}{{2a}}$

Something of import that we must not forget is the ± sign since in this way nosotros will obtain both solutions to the quadratic equation.

Example 1

Solve the equation $latex {{10}^two}+3x-4=0$.

Solution:Using $latex a=ane$, $latex b=three$ and $latex c=-4$, nosotros have:

$latex x=\frac{{-\left( 3 \right)\pm \sqrt{{{{{\left( three \right)}}^{two}}-4\left( 1 \right)\left( {-four} \correct)}}}}{{2\left( ane \right)}}$

$latex =\frac{{-three\pm \sqrt{{9+16}}}}{2}$

$latex =\frac{{-iii\pm \sqrt{{25}}}}{two}$

$latex =\frac{{-3\pm v}}{two}$

$latex =\frac{{-3-5}}{two},~~\frac{{-three+5}}{2}$

$latex =\frac{{-eight}}{2},~\frac{two}{2}=-4,~i$

Therefore, the solutions are $latex x=-4$ and $latex 10=ane$.

Case two

Solve the equation $latex 9{{10}^2}+12x+4=0$.

Solution:Using $latex a=9$, $latex b=12$ and $latex c=4$ with the quadratic formula, we have:

$latex ten=\frac{{-\left( 12 \right)\pm \sqrt{{{{{\left( 12 \right)}}^{2}}-4\left( 9 \right)\left( {4} \right)}}}}{{2\left( nine \correct)}}$

$latex =\frac{{-12\pm \sqrt{{144-144}}}}{xviii}$

$latex =\frac{{-12\pm \sqrt{{0}}}}{eighteen}$

$latex =\frac{{-12\pm 0}}{xviii}$

$latex =\frac{{-12}}{eighteen}=-\frac{{2}}{3}$

In this case, we got a zero within the square root, therefore we don't get any change from adding or subtracting nix. This means that the only solution is $latex 10=-\frac{{2}}{iii}$.

EXAMPLE 3

Solve the equation $latex 3{{x}^2}+4x+2=0$.

Solution:Using $latex a=three$, $latex b=4$ and $latex c=ii$ with the quadratic equation, we have:

$latex x=\frac{{-\left( 4 \right)\pm \sqrt{{{{{\left( 4 \correct)}}^{2}}-iv\left( three \right)\left( {two} \right)}}}}{{2\left( 3 \right)}}$

$latex =\frac{{-4\pm \sqrt{{16-24}}}}{6}$

$latex =\frac{{-4\pm \sqrt{{-8}}}}{6}$

We can come across that nosotros got a negative number inside the square root. This has no solution if we are restricted to real numbers. If y'all oasis't learned about complex numbers yet, and then we would stop here and conclude that this equation "has no solution." However, this can be solved with knowledge of complex numbers:

$latex =\frac{{-4\pm \sqrt{{-8}}}}{6}=\frac{{-four\pm 2\sqrt{{-two}}}}{6}$

$latex =\frac{{-2\pm \sqrt{{two}}i}}{3}$

$latex =-\frac{2}{three}\pm \frac{{\sqrt{2}}}{3}i$

Therefore, depending on whether we only consider real numbers or we also consider complex numbers, we have the solutions:

Simply real numbers: no solution.

Circuitous numbers:the solution is $latex =-\frac{2}{3}\pm \frac{{\sqrt{2}}}{iii}i$.

Quadratic formula – Practice issues

Solve the equation $latex -3{{x}^two}-24x-48=0$.

Choose an answer

$latex x=1$, $latex x=-2$

$latex x=-i$, $latex 10=-4$

$latex x=-four$

$latex x=-two$

Solve an equation $latex ii{{x}^2}-4x-3=0$.

Choose an answer

$latex x=0.58$, $latex 10=1.58$

$latex x=-0.58$, $latex x=2.58$

$latex ten=i.58$, $latex x=-2.58$

$latex 10=-0.58$, $latex x=0.58$

Find the value of x in the equation $latex 5{{x}^2}+6x+1=0$.

Choose an answer

$latex x=-0.2$, $latex x=-i$

$latex x=-ii$, $latex ten=1$

$latex ten=0.two$, $latex x=one$

$latex 10=two$, $latex x=-3$

See as well

Interested in learning more about algebraic expressions and factoring? Take a look at these pages:

How to Rationalize Denominators
Factoring Quadratic Polynomials
Factoring a 3rd Degree Polynomial
How to Utilise the Binomial Theorem?

Larn mathematics with our boosted resources in different topics

Acquire More