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What is the perfect square form of  x^{2}+4k+k?  

  • Option 1)

    (x+k)^{2}+4-k^{2}

  • Option 2)

    (x+k)^{2}+k^{2}-4

  • Option 3)

    (x+2)^{2}+4-k^{2}

  • Option 4)

    (x+2)^{2}+k^{2}-4

 

Answers (1)

As we have learned

Integration by perfect square method -

ax^{2}+bx+c=a\left [ x^{2}+\frac{bx}{a}+\frac{c}{a} \right ]=a\left [ \left ( x+\frac{b}{2a} \right )^{2}+\frac{c}{a}-\frac{b^{2}}{4a^{2}} \right ]

- wherein

Make the coefficient of x^{2} +ve one . 

 

 x^{2}+4x+k=x^{2}4x+4+k^{2}-4= (x+2)^{2}+(k^{2}-4)

 

 

 

 


Option 1)

(x+k)^{2}+4-k^{2}

This is incorrect

Option 2)

(x+k)^{2}+k^{2}-4

This is incorrect

Option 3)

(x+2)^{2}+4-k^{2}

This is incorrect

Option 4)

(x+2)^{2}+k^{2}-4

This is correct

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Vakul

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