Answer
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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
${x^2} + x - 6 = 0$
As we see that the highest power of x is 2 so it is a quadratic equation, if the highest power is 3 then it is a cubic equation.
Now factorize the given quadratic equation, so factorize -6 such that the sum the factors gives us 1 and product gives us negative -6.
As the coefficient of x is positive 1 and constant term is -6, so factors of -6 is such that its sum is positive 1 and product is negative 6.
So -6 can be factorize as 3 and -2,
So the sum of 3 and -2 is 1.
And the product of 3 and -2 is -6.
So the given quadratic equation is written as
$ \Rightarrow {x^2} + 3x - 2x - 6 = 0$
Now take x common from first two terms and -2 from last two terms we have,
$ \Rightarrow x\left( {x + 3} \right) - 2\left( {x + 3} \right) = 0$
Now take (x + 3) common we have,
$ \Rightarrow \left( {x + 3} \right)\left( {x - 2} \right) = 0$
Now equate both the terms to zero we have,
$ \Rightarrow \left( {x + 3} \right) = 0$, $\left( {x - 2} \right) = 0$
$ \Rightarrow x = - 3,2$
So, the roots of the quadratic equation are 2, -3.
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 = 1 and c = -6, so simply substitute these values in the given formula and simplify we will get the required answer.
Complete step by step answer:
Given equation
${x^2} + x - 6 = 0$
As we see that the highest power of x is 2 so it is a quadratic equation, if the highest power is 3 then it is a cubic equation.
Now factorize the given quadratic equation, so factorize -6 such that the sum the factors gives us 1 and product gives us negative -6.
As the coefficient of x is positive 1 and constant term is -6, so factors of -6 is such that its sum is positive 1 and product is negative 6.
So -6 can be factorize as 3 and -2,
So the sum of 3 and -2 is 1.
And the product of 3 and -2 is -6.
So the given quadratic equation is written as
$ \Rightarrow {x^2} + 3x - 2x - 6 = 0$
Now take x common from first two terms and -2 from last two terms we have,
$ \Rightarrow x\left( {x + 3} \right) - 2\left( {x + 3} \right) = 0$
Now take (x + 3) common we have,
$ \Rightarrow \left( {x + 3} \right)\left( {x - 2} \right) = 0$
Now equate both the terms to zero we have,
$ \Rightarrow \left( {x + 3} \right) = 0$, $\left( {x - 2} \right) = 0$
$ \Rightarrow x = - 3,2$
So, the roots of the quadratic equation are 2, -3.
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 = 1 and c = -6, so simply substitute these values in the given formula and simplify we will get the required answer.
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