
The equation whose roots are $\dfrac{1}{3+\sqrt{2}}$ and $\dfrac{1}{3-\sqrt{2}}$ is
( a ) $7{{x}^{2}}-6x+1=0$
( b ) $6{{x}^{2}}-6x+1=0$
( c ) ${{x}^{2}}-6x+7=0$
( d ) ${{x}^{2}}-7x+6=0$
Answer
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Hint: We are given two roots which are $\dfrac{1}{3+\sqrt{2}}$ and $\dfrac{1}{3-\sqrt{2}}$ and we have to find out the quadratic equation. As the roots are irrational and taking these roots as a and b and finding out the sum and the product of roots we are able to find out the quadratic equation.
Formula Used: a,b are the roots of the quadratic equation then the equation be in the form
${{x}^{2}}-(a+b)x+ab=0$
Complete step by step Solution:
We have to find the quadratic equation whose roots are $\dfrac{1}{3+\sqrt{2}}$ and $\dfrac{1}{3-\sqrt{2}}$
And the roots are irrational.
We know that if a, and b are the roots of the quadratic equation then the equation be in the form
${{x}^{2}}-(a+b)x+ab=0$ ………………………………………….. (1)
Here a = $\dfrac{1}{3+\sqrt{2}}$ and b = $\dfrac{1}{3-\sqrt{2}}$
a + b = $\dfrac{1}{3+\sqrt{2}}$ + $\dfrac{1}{3-\sqrt{2}}$
Then a + b = $\dfrac{6}{9-2}$ = $\dfrac{6}{7}$
And ab = $\dfrac{1}{3+\sqrt{2}}$$\times $$\dfrac{1}{3-\sqrt{2}}$
ab = $\dfrac{1}{9-2}$ = $\dfrac{1}{7}$
Now we put the value of a + b and ab in equation (1), we get
${{x}^{2}}-\dfrac{6}{7}x+\dfrac{1}{7}=0$
Solving further, we get
$7{{x}^{2}}-6x+1=0$
Hence, the equation $7{{x}^{2}}-6x+1=0$has the roots $\dfrac{1}{3+\sqrt{2}}$ and $\dfrac{1}{3-\sqrt{2}}$
Therefore, the correct option is (a).
Note: Whenever two roots a, and b are given then the quadratic equation can also be found with the help of (x – a)(x – b) =0
Then $\left( x-\dfrac{1}{3+\sqrt{2}} \right)\left( x-\dfrac{1}{3-\sqrt{2}} \right)=0$
Solving it, we get
$\left( {{x}^{2}}-\dfrac{6}{7}x+\dfrac{1}{7} \right)=0$
That is $\left( 7{{x}^{2}}-6x+1 \right)=0$
We can use any method to solve these types of questions but students must be careful in solving the roots as they do mistakes while solving the roots and are not able to get the correct answer. So students must take care while adding and multiplication of the roots
Formula Used: a,b are the roots of the quadratic equation then the equation be in the form
${{x}^{2}}-(a+b)x+ab=0$
Complete step by step Solution:
We have to find the quadratic equation whose roots are $\dfrac{1}{3+\sqrt{2}}$ and $\dfrac{1}{3-\sqrt{2}}$
And the roots are irrational.
We know that if a, and b are the roots of the quadratic equation then the equation be in the form
${{x}^{2}}-(a+b)x+ab=0$ ………………………………………….. (1)
Here a = $\dfrac{1}{3+\sqrt{2}}$ and b = $\dfrac{1}{3-\sqrt{2}}$
a + b = $\dfrac{1}{3+\sqrt{2}}$ + $\dfrac{1}{3-\sqrt{2}}$
Then a + b = $\dfrac{6}{9-2}$ = $\dfrac{6}{7}$
And ab = $\dfrac{1}{3+\sqrt{2}}$$\times $$\dfrac{1}{3-\sqrt{2}}$
ab = $\dfrac{1}{9-2}$ = $\dfrac{1}{7}$
Now we put the value of a + b and ab in equation (1), we get
${{x}^{2}}-\dfrac{6}{7}x+\dfrac{1}{7}=0$
Solving further, we get
$7{{x}^{2}}-6x+1=0$
Hence, the equation $7{{x}^{2}}-6x+1=0$has the roots $\dfrac{1}{3+\sqrt{2}}$ and $\dfrac{1}{3-\sqrt{2}}$
Therefore, the correct option is (a).
Note: Whenever two roots a, and b are given then the quadratic equation can also be found with the help of (x – a)(x – b) =0
Then $\left( x-\dfrac{1}{3+\sqrt{2}} \right)\left( x-\dfrac{1}{3-\sqrt{2}} \right)=0$
Solving it, we get
$\left( {{x}^{2}}-\dfrac{6}{7}x+\dfrac{1}{7} \right)=0$
That is $\left( 7{{x}^{2}}-6x+1 \right)=0$
We can use any method to solve these types of questions but students must be careful in solving the roots as they do mistakes while solving the roots and are not able to get the correct answer. So students must take care while adding and multiplication of the roots
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