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Question

Question: How do you solve the equation \(3({x^2} + 2) = 18\) ?...

How do you solve the equation 3(x2+2)=183({x^2} + 2) = 18 ?

Explanation

Solution

At first, we will divide both sides of the equation by 33 . Then we will get a simplified form of the equation and the equation will be a quadratic equation. Then if the equation will be in the form x2=a2{x^2} = {a^2} the solution will be x=±ax = \pm a . If the equation will be in the form x2=b{x^2} = b . The solution will be x=±bx = \pm \sqrt b .

Complete step by step answer:
We have given;
3(x2+2)=183({x^2} + 2) = 18
At first, we will divide both sides of the equation by 33 .
We will get;
x2+2=183\Rightarrow {x^2} + 2 = \dfrac{{18}}{3}
Simplifying the above equation we get;
x2+2=6\Rightarrow {x^2} + 2 = 6
Subtracting 22 from both side we get;
x2=4\Rightarrow {x^2} = 4
We know that 4=224 = {2^2} . Applying this in the above equation we get;
x2=22\Rightarrow {x^2} = {2^2}
Now we know that if x2=a2{x^2} = {a^2} then the solution will be;
x=±ax = \pm a
Applying this in the above equation we get;
x=±2\Rightarrow x = \pm 2
So the required solution is x=±2x = \pm 2 .
Alternative Method:
We have given;
3(x2+2)=183({x^2} + 2) = 18
At first, we will multiply the left-hand side and get;
3x2+6=18\Rightarrow 3{x^2} + 6 = 18
Subtracting 66 from both side of the above equation we get;
3x2=12\Rightarrow 3{x^2} = 12
Dividing 33 from both side we get;
x2=4\Rightarrow {x^2} = 4
We know that 4=224 = {2^2} . Applying this in the above equation we get;
x2=22\Rightarrow {x^2} = {2^2}
Now we know that if x2=a2{x^2} = {a^2} then the solution will be;
x=±ax = \pm a
Applying this in the above equation we get;
x=±2\Rightarrow x = \pm 2

So the required solution is x=±2x = \pm 2 .

Note: The general quadratic equation is in the form ax2+bx+c=0a{x^2} + bx + c = 0 . After solving this question 3(x2+2)=183({x^2} + 2) = 18 we get the equation x2=4{x^2} = 4 . Comparing this equation with the general form of the quadratic equation we get a=1a = 1 , the value of bb is zero that’s why there is no term with xx and cc is equal to 4- 4 . Students remember the solution of the quadratic equation always gives two values.