Question

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Find the roots of the Quadratic equation given above.
Kindly solve fast.​

Answer 1

Answer:

x= 42k^2x +√(252k^4+504)/28

x= 42k^2x -√(252k^4+504)/28

Step-by-step explanation:

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Answer 2

Ans:

two roots are :

$x = \frac{ 42 {k}^{2} \ + \sqrt{ 1764 {k}^{2} + 504}}{ 28 } \: \: \: \\ \\ x = \frac{ 42 {k}^{2} \ - \sqrt{ 1764 {k}^{2} + 504}}{ 28 }$

Explanation:

comparing above equation with ax²+bx+c=0, weget , a= 14 , b=-42k², c=-9

root x:

$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }$

$x = \frac{ -(-42 {k}^{2} ) \pm \sqrt{(-42 {k}^{2} )^2 - 4(14)(-9)}}{ 2(14) }$

$x = \frac{ 42 {k}^{2} \pm \sqrt{ 1764 {k}^{2} + 504}}{ 28 }$

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