Question

Find the roots of quadratic equation if exists

Answer 1

Answer:

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Answer 2

For the equation 2x

For the equation 2x 2

For the equation 2x 2 −2

For the equation 2x 2 −2 2

For the equation 2x 2 −2 2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots,

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 )

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1 =8−8=0

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1 =8−8=0We got Δ=0

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1 =8−8=0We got Δ=0∴ Root for the equation exists and they are equal.

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1 =8−8=0We got Δ=0∴ Root for the equation exists and they are equal.Let the roots of the equation be x

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1 =8−8=0We got Δ=0∴ Root for the equation exists and they are equal.Let the roots of the equation be x 1

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1 =8−8=0We got Δ=0∴ Root for the equation exists and they are equal.Let the roots of the equation be x 1

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1 =8−8=0We got Δ=0∴ Root for the equation exists and they are equal.Let the roots of the equation be x 1 and x

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1 =8−8=0We got Δ=0∴ Root for the equation exists and they are equal.Let the roots of the equation be x 1 and x 2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1 =8−8=0We got Δ=0∴ Root for the equation exists and they are equal.Let the roots of the equation be x 1 and x 2

For the equation 2x 2 −2 2 x+1=0; a=2, b=−2 2 , c=1By Shridhar acharya's formula to test the existence of the roots, we have b 2 −4ac=(2 2 ) 2 −4×2×1 =8−8=0We got Δ=0∴ Root for the equation exists and they are equal.Let the roots of the equation be x 1 and x 2