Answer
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Hint: At first, we will divide both sides of the equation by $3$ . 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 ${x^2} = {a^2}$ the solution will be $x = \pm a$ . If the equation will be in the form ${x^2} = b$ . The solution will be $x = \pm \sqrt b$ .
Complete step by step answer:
We have given;
$3({x^2} + 2) = 18$
At first, we will divide both sides of the equation by $3$ .
We will get;
$\Rightarrow {x^2} + 2 = \dfrac{{18}}{3}$
Simplifying the above equation we get;
$\Rightarrow {x^2} + 2 = 6$
Subtracting $2$ from both side we get;
$\Rightarrow {x^2} = 4$
We know that $4 = {2^2}$ . Applying this in the above equation we get;
$\Rightarrow {x^2} = {2^2}$
Now we know that if ${x^2} = {a^2}$ then the solution will be;
$x = \pm a$
Applying this in the above equation we get;
$\Rightarrow x = \pm 2$
So the required solution is $x = \pm 2$ .
Alternative Method:
We have given;
$3({x^2} + 2) = 18$
At first, we will multiply the left-hand side and get;
$\Rightarrow 3{x^2} + 6 = 18$
Subtracting $6$ from both side of the above equation we get;
$\Rightarrow 3{x^2} = 12$
Dividing $3$ from both side we get;
$\Rightarrow {x^2} = 4$
We know that $4 = {2^2}$ . Applying this in the above equation we get;
$\Rightarrow {x^2} = {2^2}$
Now we know that if ${x^2} = {a^2}$ then the solution will be;
$x = \pm a$
Applying this in the above equation we get;
$\Rightarrow x = \pm 2$
So the required solution is $x = \pm 2$ .
Note: The general quadratic equation is in the form $a{x^2} + bx + c = 0$ . After solving this question $3({x^2} + 2) = 18$ we get the equation ${x^2} = 4$ . Comparing this equation with the general form of the quadratic equation we get $a = 1$ , the value of $b$ is zero that’s why there is no term with $x$ and $c$ is equal to $- 4$ . Students remember the solution of the quadratic equation always gives two values.
Complete step by step answer:
We have given;
$3({x^2} + 2) = 18$
At first, we will divide both sides of the equation by $3$ .
We will get;
$\Rightarrow {x^2} + 2 = \dfrac{{18}}{3}$
Simplifying the above equation we get;
$\Rightarrow {x^2} + 2 = 6$
Subtracting $2$ from both side we get;
$\Rightarrow {x^2} = 4$
We know that $4 = {2^2}$ . Applying this in the above equation we get;
$\Rightarrow {x^2} = {2^2}$
Now we know that if ${x^2} = {a^2}$ then the solution will be;
$x = \pm a$
Applying this in the above equation we get;
$\Rightarrow x = \pm 2$
So the required solution is $x = \pm 2$ .
Alternative Method:
We have given;
$3({x^2} + 2) = 18$
At first, we will multiply the left-hand side and get;
$\Rightarrow 3{x^2} + 6 = 18$
Subtracting $6$ from both side of the above equation we get;
$\Rightarrow 3{x^2} = 12$
Dividing $3$ from both side we get;
$\Rightarrow {x^2} = 4$
We know that $4 = {2^2}$ . Applying this in the above equation we get;
$\Rightarrow {x^2} = {2^2}$
Now we know that if ${x^2} = {a^2}$ then the solution will be;
$x = \pm a$
Applying this in the above equation we get;
$\Rightarrow x = \pm 2$
So the required solution is $x = \pm 2$ .
Note: The general quadratic equation is in the form $a{x^2} + bx + c = 0$ . After solving this question $3({x^2} + 2) = 18$ we get the equation ${x^2} = 4$ . Comparing this equation with the general form of the quadratic equation we get $a = 1$ , the value of $b$ is zero that’s why there is no term with $x$ and $c$ is equal to $- 4$ . Students remember the solution of the quadratic equation always gives two values.
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