Answer
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Hint: Completing the Square is a method that may be used for any quadratic equation. By adjusting your constant, you can create a perfect square on the left side of the equation. A perfect square can be factored into two identical binomials, which you can use to solve for any valid values of x.
Complete step-by-step answer:
Suppose $a{{x}^{2}}+bx+c=0$ is the given quadratic equation. Then follow the given steps to solve it by completing the square method.
Write the equation in the form, such that c is on the right side.
If a is not equal to 1, then divide the complete equation by a, such that the coefficient of ${{x}^{2}}$ is 1.
Now add the square of half of the coefficient of term-x, ${{\left( \dfrac{b}{2a} \right)}^{2}}$on both sides.
Factorize the left side of the equation as the square of the binomial term.
Take the square root on both the sides
Solve for variable x and find the roots.
Let us consider the quadratic equation,
$5{{x}^{2}}-6x-2=0$
Dividing both sides by 5, we get
\[{{x}^{2}}-\left( \dfrac{6}{5} \right)x-\left( \dfrac{2}{5} \right)=0\]
Add and subtract the term ${{\left( \dfrac{3}{5} \right)}^{2}}$ on the left side of the equation, we get
\[{{x}^{2}}-\left( \dfrac{6}{5} \right)x+{{\left( \dfrac{3}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}-\left( \dfrac{2}{5} \right)=0\]
Rearranging the terms, we get
\[{{x}^{2}}-2\left( \dfrac{3}{5} \right)x+{{\left( \dfrac{3}{5} \right)}^{2}}=\dfrac{9}{25}+\dfrac{2}{5}\]
\[{{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{19}{25}\]
Squaring roots on the both sides, we get
\[\left( x-\dfrac{3}{5} \right)=\pm \sqrt{\dfrac{19}{25}}\]
\[x=\dfrac{3}{5}\pm \dfrac{\sqrt{19}}{5}\]
$x=\dfrac{3\pm \sqrt{19}}{5}$
Therefore, the correct option is (d).
Note: Completing the square method is one of the methods to find the roots of the given quadratic equation. A polynomial equation with degree equal to two is known as a quadratic equation. Quad means four but Quadratic means to make square.
Complete step-by-step answer:
Suppose $a{{x}^{2}}+bx+c=0$ is the given quadratic equation. Then follow the given steps to solve it by completing the square method.
Write the equation in the form, such that c is on the right side.
If a is not equal to 1, then divide the complete equation by a, such that the coefficient of ${{x}^{2}}$ is 1.
Now add the square of half of the coefficient of term-x, ${{\left( \dfrac{b}{2a} \right)}^{2}}$on both sides.
Factorize the left side of the equation as the square of the binomial term.
Take the square root on both the sides
Solve for variable x and find the roots.
Let us consider the quadratic equation,
$5{{x}^{2}}-6x-2=0$
Dividing both sides by 5, we get
\[{{x}^{2}}-\left( \dfrac{6}{5} \right)x-\left( \dfrac{2}{5} \right)=0\]
Add and subtract the term ${{\left( \dfrac{3}{5} \right)}^{2}}$ on the left side of the equation, we get
\[{{x}^{2}}-\left( \dfrac{6}{5} \right)x+{{\left( \dfrac{3}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}-\left( \dfrac{2}{5} \right)=0\]
Rearranging the terms, we get
\[{{x}^{2}}-2\left( \dfrac{3}{5} \right)x+{{\left( \dfrac{3}{5} \right)}^{2}}=\dfrac{9}{25}+\dfrac{2}{5}\]
\[{{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{19}{25}\]
Squaring roots on the both sides, we get
\[\left( x-\dfrac{3}{5} \right)=\pm \sqrt{\dfrac{19}{25}}\]
\[x=\dfrac{3}{5}\pm \dfrac{\sqrt{19}}{5}\]
$x=\dfrac{3\pm \sqrt{19}}{5}$
Therefore, the correct option is (d).
Note: Completing the square method is one of the methods to find the roots of the given quadratic equation. A polynomial equation with degree equal to two is known as a quadratic equation. Quad means four but Quadratic means to make square.
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