
How do you solve the equation $9{{x}^{2}}+30x+25=11$ by completing the square?
Answer
528k+ views
Hint: From the question we have been asked to find the solution to the equation $9{{x}^{2}}+30x+25=11$ using the completing square method. To solve this question first, we have to eliminate $9$ from the coefficient of ${{x}^{2}}$ and see whether we got the coefficient as $1$ and next we will add required integer or fraction on both sides of equation and solve the given question in the form of completing square method form.
Complete step by step solution:
From the question given we have to solve the $9{{x}^{2}}+30x+25=11$ by using the complete square method.
Shift $11$ from the right-hand side equation to the left-hand side of the equation,
By shifting we will get,
$\Rightarrow 9{{x}^{2}}+30x+25-11=0$
$\Rightarrow 9{{x}^{2}}+30x+14=0$
So, in the process of solving the question first we will remove the $9$ before ${{x}^{2}}$ by dividing $9$ on both sides of the equation.
So, the equation will be reduced as follows.
$\Rightarrow {{x}^{2}}+\dfrac{30}{9}x+\dfrac{14}{9}=0$
$\Rightarrow {{x}^{2}}+\dfrac{10}{3}x+\dfrac{14}{9}=0$
Here we must make sure that the coefficient of ${{x}^{2}}$ should be 1 and nothing other than that, from the above equation we can clearly see that its coefficient is 1. So, we will further continue our solution as follows.
Here we will add and subtract the integer $\dfrac{25}{9}$ to the equation for the simplification. So, the equation will be reduced as follows.
$\Rightarrow {{x}^{2}}+\dfrac{10}{3}x+\dfrac{14}{9}+\dfrac{25}{9}-\dfrac{25}{9}=0$
$\Rightarrow {{x}^{2}}+\dfrac{10}{3}x+\dfrac{25}{9}-\dfrac{11}{9}=0$
Here now we will bring the integer $\dfrac{11}{9}$ to the right-hand side of the equation. So, the equation will be reduced as follows.
$\Rightarrow {{x}^{2}}+\dfrac{10}{3}x+\dfrac{25}{9}=\dfrac{11}{9}$
Now, here we can clearly see that the expression on the left-hand side of the equation is a complete square. So, after rewriting the equation we reduced the equation as follows.
$\Rightarrow {{\left( x+\dfrac{5}{3} \right)}^{2}}=\dfrac{11}{9}$
Therefore, in this way we complete the square $ {{\left( x+\dfrac{5}{3} \right)}^{2}}=\dfrac{11}{9}$ to solve the question.
The solution for the equation will be as follows.
$\Rightarrow {{\left( x+\dfrac{5}{3} \right)}^{2}}=\dfrac{11}{9}$
Here we will shift the square of the left-hand side to the right-hand side then it will be square root to the right-hand side equation.
$\Rightarrow \left( x+\dfrac{5}{3} \right)=\pm\sqrt{\dfrac{11}{9}}$
$\Rightarrow \left( x+\dfrac{5}{3} \right)=\pm\dfrac{\sqrt{11}}{3}$
$\Rightarrow x=\pm\dfrac{\sqrt{11}}{3}-\dfrac{5}{3}$
$\Rightarrow x=\pm\dfrac{\sqrt{11}-5}{3}$
Therefore, the solution for the question using complete square method will be $x=\pm\dfrac{\sqrt{11}-5}{3}$ and in this way we complete the square $9{{x}^{2}}+30x+25=11$ to solve the question.
Note: Students should be very careful while transforming the equation into the square if any error comes in that step the total solution will be wrong. Students should know that while square root is cancelled there will be two possible values one is positive and another one is negative this point should be kept in mind. Students should always apply this method only when the coefficient of ${{x}^{2}}$ is 1.
Complete step by step solution:
From the question given we have to solve the $9{{x}^{2}}+30x+25=11$ by using the complete square method.
Shift $11$ from the right-hand side equation to the left-hand side of the equation,
By shifting we will get,
$\Rightarrow 9{{x}^{2}}+30x+25-11=0$
$\Rightarrow 9{{x}^{2}}+30x+14=0$
So, in the process of solving the question first we will remove the $9$ before ${{x}^{2}}$ by dividing $9$ on both sides of the equation.
So, the equation will be reduced as follows.
$\Rightarrow {{x}^{2}}+\dfrac{30}{9}x+\dfrac{14}{9}=0$
$\Rightarrow {{x}^{2}}+\dfrac{10}{3}x+\dfrac{14}{9}=0$
Here we must make sure that the coefficient of ${{x}^{2}}$ should be 1 and nothing other than that, from the above equation we can clearly see that its coefficient is 1. So, we will further continue our solution as follows.
Here we will add and subtract the integer $\dfrac{25}{9}$ to the equation for the simplification. So, the equation will be reduced as follows.
$\Rightarrow {{x}^{2}}+\dfrac{10}{3}x+\dfrac{14}{9}+\dfrac{25}{9}-\dfrac{25}{9}=0$
$\Rightarrow {{x}^{2}}+\dfrac{10}{3}x+\dfrac{25}{9}-\dfrac{11}{9}=0$
Here now we will bring the integer $\dfrac{11}{9}$ to the right-hand side of the equation. So, the equation will be reduced as follows.
$\Rightarrow {{x}^{2}}+\dfrac{10}{3}x+\dfrac{25}{9}=\dfrac{11}{9}$
Now, here we can clearly see that the expression on the left-hand side of the equation is a complete square. So, after rewriting the equation we reduced the equation as follows.
$\Rightarrow {{\left( x+\dfrac{5}{3} \right)}^{2}}=\dfrac{11}{9}$
Therefore, in this way we complete the square $ {{\left( x+\dfrac{5}{3} \right)}^{2}}=\dfrac{11}{9}$ to solve the question.
The solution for the equation will be as follows.
$\Rightarrow {{\left( x+\dfrac{5}{3} \right)}^{2}}=\dfrac{11}{9}$
Here we will shift the square of the left-hand side to the right-hand side then it will be square root to the right-hand side equation.
$\Rightarrow \left( x+\dfrac{5}{3} \right)=\pm\sqrt{\dfrac{11}{9}}$
$\Rightarrow \left( x+\dfrac{5}{3} \right)=\pm\dfrac{\sqrt{11}}{3}$
$\Rightarrow x=\pm\dfrac{\sqrt{11}}{3}-\dfrac{5}{3}$
$\Rightarrow x=\pm\dfrac{\sqrt{11}-5}{3}$
Therefore, the solution for the question using complete square method will be $x=\pm\dfrac{\sqrt{11}-5}{3}$ and in this way we complete the square $9{{x}^{2}}+30x+25=11$ to solve the question.
Note: Students should be very careful while transforming the equation into the square if any error comes in that step the total solution will be wrong. Students should know that while square root is cancelled there will be two possible values one is positive and another one is negative this point should be kept in mind. Students should always apply this method only when the coefficient of ${{x}^{2}}$ is 1.
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