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If the equation (a-2)x^{2}-(a-4)x-2=0 has a difference of roots as 3, then the value of a is

  • Option 1)

    1,3

  • Option 2)

    3,\frac{3}{2}

  • Option 3)

    2,\frac{3}{2}

  • Option 4)

    \frac{3}{2},1

 

Answers (1)

As we learnt in 

Sum of Roots in Quadratic Equation -

\alpha +\beta = \frac{-b}{a}

- wherein

\alpha \: and\beta are root of quadratic equation

ax^{2}+bx+c=0

a,b,c\in C

 

 

To form a Quadratic Equation given the roots -

x^{2}-Sx+P= 0

- wherein

S = Sum of roots

P = Product of roots

 

 \left ( a-2 \right )n^{2}-\left ( a-4 \right )x-2=0

Difference of roots

  =\frac{\sqrt{\left ( a-4 \right )^{2}-4\times \left ( -2 \right )\left ( a-2 \right )}}{\left ( a-2 \right )}=3\\*\\* \sqrt{\left ( a-4 \right )^{2}+8\left ( a-2 \right )}=3\left ( a-2 \right )\\*\\*\sqrt{a^{2}-8a+16+8a-16}=3\left ( a-2 \right )\\*\\* \left | a \right |=3\left ( a-2 \right )

a=3a-6 or -a=3a-6

a=3   or   a=3/2


Option 1)

1,3

Incorrect

Option 2)

3,\frac{3}{2}

Correct

Option 3)

2,\frac{3}{2}

Incorrect

Option 4)

\frac{3}{2},1

Incorrect

Posted by

Vakul

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