How do you solve the equation $\dfrac{2}{3}{x^2} - 4 = 12?$
Answer
Verified
438k+ views
Hint: First of all take the given expression, take all the constants on one side and the term with variable on one side and then simplify the equation for the resultant required value for the unknown term “x”.
Complete step-by-step solution:
Take the given expression: $\dfrac{2}{3}{x^2} - 4 = 12$
Move constant on the right hand side of the equation. When you move any term from one side to another then the sign of the term also changes. Positive term becomes negative and negative term becomes positive.
$\dfrac{2}{3}{x^2} = 12 + 4$
Simplify the expression on the right hand side of the equation.
$\dfrac{2}{3}{x^2} = 16$
The term in division on one side, if moved to the opposite side then it goes to the numerator part.
$2{x^2} = 16 \times 3$
Simplify the above equation-
$2{x^2} = 48$
The term multiplicative on one side if moved to the opposite side then it goes to the denominator part.
$ \Rightarrow {x^2} = \dfrac{{48}}{2}$
Find factors of the term on the numerator.
$ \Rightarrow {x^2} = \dfrac{{24 \times 2}}{2}$
Common factors from the numerator and the denominator cancels each other. Therefore, remove from the numerator and the denominator.
$ \Rightarrow {x^2} = 24$
Take square root on both sides of the above equation.
$ \Rightarrow \sqrt {{x^2}} = \sqrt {24} $
Square and square root cancel each other on the left hand side of the equation.
$
\Rightarrow x = \sqrt {4 \times 6} \\
\Rightarrow x = \sqrt {{2^2} \times 6} \\
\Rightarrow x = \sqrt {{2^2}} \times \sqrt 6 \\
$
Square and square root cancel in the above equation.
$ \Rightarrow x = \pm 2\sqrt 6 $
This is the required solution.
Note: Always remember that the square of positive or the negative number always gives us the positive number but the square root of positive number gives positive or the negative number therefore, we have kept plus or minus ahead of the number. Be careful about the plus or minus signs. Be good in multiples and remember at least twenty.
Complete step-by-step solution:
Take the given expression: $\dfrac{2}{3}{x^2} - 4 = 12$
Move constant on the right hand side of the equation. When you move any term from one side to another then the sign of the term also changes. Positive term becomes negative and negative term becomes positive.
$\dfrac{2}{3}{x^2} = 12 + 4$
Simplify the expression on the right hand side of the equation.
$\dfrac{2}{3}{x^2} = 16$
The term in division on one side, if moved to the opposite side then it goes to the numerator part.
$2{x^2} = 16 \times 3$
Simplify the above equation-
$2{x^2} = 48$
The term multiplicative on one side if moved to the opposite side then it goes to the denominator part.
$ \Rightarrow {x^2} = \dfrac{{48}}{2}$
Find factors of the term on the numerator.
$ \Rightarrow {x^2} = \dfrac{{24 \times 2}}{2}$
Common factors from the numerator and the denominator cancels each other. Therefore, remove from the numerator and the denominator.
$ \Rightarrow {x^2} = 24$
Take square root on both sides of the above equation.
$ \Rightarrow \sqrt {{x^2}} = \sqrt {24} $
Square and square root cancel each other on the left hand side of the equation.
$
\Rightarrow x = \sqrt {4 \times 6} \\
\Rightarrow x = \sqrt {{2^2} \times 6} \\
\Rightarrow x = \sqrt {{2^2}} \times \sqrt 6 \\
$
Square and square root cancel in the above equation.
$ \Rightarrow x = \pm 2\sqrt 6 $
This is the required solution.
Note: Always remember that the square of positive or the negative number always gives us the positive number but the square root of positive number gives positive or the negative number therefore, we have kept plus or minus ahead of the number. Be careful about the plus or minus signs. Be good in multiples and remember at least twenty.
Recently Updated Pages
A uniform rod of length l and mass m is free to rotate class 10 physics CBSE
Solve the following pairs of linear equations by elimination class 10 maths CBSE
What could be the possible ones digits of the square class 10 maths CBSE
Where was the Great Bath found A Harappa B Mohenjodaro class 10 social science CBSE
PQ is a tangent to a circle with centre O at the point class 10 maths CBSE
The measures of two adjacent sides of a parallelogram class 10 maths CBSE
Trending doubts
Imagine that you have the opportunity to interview class 10 english CBSE
Find the area of the minor segment of a circle of radius class 10 maths CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
Frogs can live both on land and in water name the adaptations class 10 biology CBSE
Fill in the blank One of the students absent yesterday class 10 english CBSE
Write a letter to the Principal of your school requesting class 10 english CBSE