Answer
Verified
389.7k+ views
Hint: In this particular question use the concept of factorization, the factorization is done on the basis of coefficient of x, so we have to factorize the multiplication of constant term and coefficient of ${x^2}$ such that its sum gives us the coefficient of x and product gives us the multiplication of coefficient of ${x^2}$ and the constant term so use these concepts to reach the solution of the question.
Complete step by step answer:
Given equation
$6{x^2} - 5x - 6$
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 the multiplication of constant term and coefficient of ${x^2}$ such that the sum or difference gives us -5 and product gives us $\left( {6 \times - 6} \right) = - 36$
As the coefficient of ${x^2}$ is 6, coefficient of x is negative 5 and constant term is -6, so factors of multiplication of constant term and coefficient of ${x^2}$ i.e. $\left( {6 \times - 6} \right) = - 36$, such that the sum of the factors will give us the coefficient of x i.e. -5 and the product of the factors will give us $\left( {6 \times - 6} \right) = - 36$
So -36 can be factorize as -9 and 4,
So the sum of -9 and 4 is -5.
And the product of -9 and 4 is -36.
So the given quadratic equation is written as
$ \Rightarrow 6{x^2} - 9x + 4x - 6$
Now take 3x common from first two terms and 2 from last two terms we have,
$ \Rightarrow 3x\left( {2x - 3} \right) + 2\left( {2x - 3} \right)$
Now take (2x – 3) common we have,
$ \Rightarrow \left( {2x - 3} \right)\left( {3x + 2} \right)$
So this is the required factorization 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 = 6, b = -5 and c= -6, so simply substitute these values in the given equation and simplify we will get the values of x let the values of x be p and q, so the factors are (x – p) and (x – q) respectively.
Complete step by step answer:
Given equation
$6{x^2} - 5x - 6$
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 the multiplication of constant term and coefficient of ${x^2}$ such that the sum or difference gives us -5 and product gives us $\left( {6 \times - 6} \right) = - 36$
As the coefficient of ${x^2}$ is 6, coefficient of x is negative 5 and constant term is -6, so factors of multiplication of constant term and coefficient of ${x^2}$ i.e. $\left( {6 \times - 6} \right) = - 36$, such that the sum of the factors will give us the coefficient of x i.e. -5 and the product of the factors will give us $\left( {6 \times - 6} \right) = - 36$
So -36 can be factorize as -9 and 4,
So the sum of -9 and 4 is -5.
And the product of -9 and 4 is -36.
So the given quadratic equation is written as
$ \Rightarrow 6{x^2} - 9x + 4x - 6$
Now take 3x common from first two terms and 2 from last two terms we have,
$ \Rightarrow 3x\left( {2x - 3} \right) + 2\left( {2x - 3} \right)$
Now take (2x – 3) common we have,
$ \Rightarrow \left( {2x - 3} \right)\left( {3x + 2} \right)$
So this is the required factorization 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 = 6, b = -5 and c= -6, so simply substitute these values in the given equation and simplify we will get the values of x let the values of x be p and q, so the factors are (x – p) and (x – q) respectively.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE