Question

How do you solve \[{x^2} + 6x - 16 = 0\]?

Answer 1

Hint:In order to determine solution of the above quadratic question use the Splitting up the middle method by first multiplying the coefficient of ${x^2}$ with the constant term and factorise it into two factors such that either addition or subtraction gives us the middle term and the product of the same gives us back the multiplication we have calculated. Pull out common from the first two terms and the last two terms and then again pull out the binomial parenthesis .Put all the things equal to zero to obtain the required solution.

Complete step by step solution:
Given a quadratic equation, \[{x^2} + 6x - 16 = 0\] let it be $f(x)$
$f(x) = {x^2} + 6x - 16 = 0$
Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 1
b becomes 6
And c becomes -16
To find the quadratic factorization we’ll use splitting up the middle term method
So first calculate the product of coefficient of ${x^2}$and the constant term which comes to be
$ = ( - 16) \times 1 = - 16$
Now second Step is to the find the 2 factors of the number 2 such that the whether addition or
subtraction of those numbers is equal to the middle term or coefficient of x and the product of those
factors result in the value of constant .
So if we factorize $ - 16$, the answer comes to be 8 and -2 as $8 - 2 = 6$ that is the middle term and \[ - 2 \times 8 = - 16\] which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained ,so equation $f(x)$ becomes
$f(x) = {x^2} + 8x - 2x - 16 = 0$
Now taking common from the first 2 terms and last 2 terms
$f(x) = x(x + 8) - 2(x + 8) = 0$
Finding the common binomial parenthesis, the equation becomes
\[
f(x) = (x + 8)(x - 2) = 0 \\
x + 8 = 0 \\
\Rightarrow x = - 8 \\
x - 2 = 0 \\
\Rightarrow x = 2 \\
\]
$x = 2, - 8$
Hence , We have successfully obtained the solution to the equation as $x = 2, - 8$
Alternative:
You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
x1,x2 are root or solutions to quadratic equation $a{x^2} + bx + c$
Hence the factors will be $(x - x1)\,and\,(x - x2)\,$.

Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.
3. Write the factors when the middle term is split using the proper sign.