Question

The quadratic equations having the roots $\dfrac{1}{\left[ 10-\sqrt{72} \right]}$ and $\dfrac{1}{\left[ 10+6\sqrt{2} \right]}$ is
$1)28{{x}^{2}}-20x+1=0$
$2)20{{x}^{2}}-28x+1=0$
$3){{x}^{2}}-20x+28=0$
$4){{x}^{2}}-28x+20=0$

Answer 1

Hint: To solve this question we need to have knowledge of quadratic equations. Quadratic equation is basically a type of equation whose highest power is $2$. The quadratic equation in general form is written as $a{{x}^{2}}+bx+c=0$ . If the roots given to us are $p$ and $q$ then the quadratic equation becomes ${{x}^{2}}-\left( p+q \right)x+pq=0$ . To solve the question we will find out the sum of the roots and product of the roots and can substitute in the above form of quadratic equation.

Complete step by step answer:
The question ask to find the quadratic equation when the two of the roots given are $\dfrac{1}{\left[ 10-\sqrt{72} \right]}$ and $\dfrac{1}{\left[ 10+6\sqrt{2} \right]}$ . Firstly we will write the quadratic equation $a{{x}^{2}}+bx+c=0$ in form of the sum of the roots and product of the roots. The equation in this form is:
$\Rightarrow {{x}^{2}}-\left( \text{sum of roots} \right)x+\text{product of roots=0}$
 The first step will be to find the sum of roots for that will be adding to have the two values given to us which are $\dfrac{1}{\left[ 10-\sqrt{72} \right]}$ and $\dfrac{1}{\left[ 10+6\sqrt{2} \right]}$. On addition we will get the following expression:
$\Rightarrow ~\dfrac{1}{\left[ 10-\sqrt{72} \right]}+\dfrac{1}{\left[ 10+6\sqrt{2} \right]}$
In the above expression we see that $\sqrt{72}$ could be written as $6\sqrt{2}$. On substituting in the place of $\sqrt{72}$ we get:
$\Rightarrow ~\dfrac{1}{\left[ 10-6\sqrt{2} \right]}+\dfrac{1}{\left[ 10+6\sqrt{2} \right]}$
We will find the L.C.M of the above expression. On doing so we get:
$\Rightarrow ~\dfrac{10+6\sqrt{2}+10-6\sqrt{2}}{\left[ 10-6\sqrt{2} \right]\left[ 10+6\sqrt{2} \right]}$
We see that the denominator are conjugate to each other. So we will apply the formula $(a-b)(a+b)={{a}^{2}}-{{b}^{2}}$ . On applying the formula we get:
$\Rightarrow ~\dfrac{20}{\left[ {{10}^{2}}-6{{\sqrt{2}}^{2}} \right]}$
$\Rightarrow ~\dfrac{20}{\left[ 100-72 \right]}$
$\Rightarrow ~\dfrac{20}{28}$
 The next step is to find the product of the roots. On doing so we get:
$\Rightarrow ~\dfrac{1}{\left[ 10-6\sqrt{2} \right]}\times \dfrac{1}{\left[ 10+6\sqrt{2} \right]}$
$\Rightarrow ~\dfrac{1}{\left[ {{10}^{2}}-6{{\sqrt{2}}^{2}} \right]}$
$\Rightarrow ~\dfrac{1}{28}$
Now we will substitute the values in the equation
$\Rightarrow {{x}^{2}}-\left( \text{sum of roots} \right)x+\text{product of roots=0}$
On substituting the values we get:
$\Rightarrow {{x}^{2}}-\dfrac{20}{28}x+\dfrac{1}{28}\text{=0}$
$\Rightarrow \dfrac{28}{28}{{x}^{2}}-\dfrac{20}{28}x+\dfrac{1}{28}\text{=0}$
$\Rightarrow 28{{x}^{2}}-20x+1\text{=0}$
$\therefore $ The quadratic equations having the roots $\dfrac{1}{\left[ 10-\sqrt{72} \right]}$ and $\dfrac{1}{\left[ 10+6\sqrt{2} \right]}$ is $28{{x}^{2}}-20x+1=0$

So, the correct answer is “Option 1”.

Note: In these type of quadratic equation when the roots given to us do remember the formula used in terms of sum and product is ${{x}^{2}}-\left( \text{sum of roots} \right)x+\text{product of roots=0}$. We can check the solution whether it is correct or not by finding the roots of the quadratic equation $28{{x}^{2}}-20x+1\text{=0}$, with the help of the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .