Question

Solve the quadratic equation and find values of $x$: ${x^2} + 6x - 16 = 0$

Answer 1

Hint: In this particular question use the concept of factorization, the factorization is done on the basis of coefficient of x and the constant term, so we have to factorize the constant term such that its sum gives us the coefficient of x and product gives us the constant term so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation
${x^2} + 6x - 16 = 0$
As we see that the highest power of x is 2 so it is a quadratic equation, if the highest power was 3 then it is a cubic equation.
Now factorize the given quadratic equation, so factorize 16 such that the sum or difference gives us 6 and product gives us positive 16 or negative 16.
As the coefficient of x is positive 6 and constant term is -16, so factors of 16 is such that its sum is positive 6 and product is negative 16
So 16 can be factorize as 8 and -2,
So the sum of 8 and -2 is 6.
And the product of 8 and -2 is -16.
So the given quadratic equation is written as
$ \Rightarrow {x^2} + 8x - 2x - 16 = 0$
Now take x common from first two terms and -2 from last two terms we have,
$ \Rightarrow x\left( {x + 8} \right) - 2\left( {x + 8} \right) = 0$
Now take (x + 8) common we have,
$ \Rightarrow \left( {x + 8} \right)\left( {x - 2} \right) = 0$
Now equate both the terms to zero we have,
$ \Rightarrow \left( {x + 8} \right) = 0$, $\left( {x - 2} \right) = 0$
$ \Rightarrow x = - 8,2$
So this is the required solution of the given quadratic equation.

Note: We can also solve this quadratic equation directly by using the quadratic formula which is given as, $\left( {x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right)$, where a, b and c are the coefficients of ${x^2}$, x and constant term respectively, so on comparing, a = 1, b = 6 and c= -16, so simply substitute these values in the given equation and simplify we will get the required answer.