Find the roots of x^2 - 10x + 9 using square method
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Answered by
2
x² -10x + 9 =0
x² -9x - x + 9 =0
x(x - 9) - 1 ( x - 9) =0
(x - 9)(x-1) =0.
x - 9 = 0
x = 9
and x - 1 = 0
x = 1
x² -9x - x + 9 =0
x(x - 9) - 1 ( x - 9) =0
(x - 9)(x-1) =0.
x - 9 = 0
x = 9
and x - 1 = 0
x = 1
Answered by
3
Hey Mate!
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Given,
![{x}^{2} - 10x + 9 = 0 \\ \\ {x}^{2} - 10x = - 9 \\ \\ add \: {5}^{2} on \: both \: sides \\ \\ {x}^{2} - 2x(5) + {5}^{2} = - 9 + {5}^{2} \\ \\ {(x - 5)}^{2} = - 9 + 25 \\ \\ {(x - 5)}^{2} = 16 \\ \\ x - 5 = + and \: - \sqrt{16} \\ \\ x - 5 = + and \: - 4 \\ \\ x = 4 + 5 \: \: \: x = - 4 + 5 \\ \\ x = 9 \: \: \: \: \: \: \: \: x = 1 {x}^{2} - 10x + 9 = 0 \\ \\ {x}^{2} - 10x = - 9 \\ \\ add \: {5}^{2} on \: both \: sides \\ \\ {x}^{2} - 2x(5) + {5}^{2} = - 9 + {5}^{2} \\ \\ {(x - 5)}^{2} = - 9 + 25 \\ \\ {(x - 5)}^{2} = 16 \\ \\ x - 5 = + and \: - \sqrt{16} \\ \\ x - 5 = + and \: - 4 \\ \\ x = 4 + 5 \: \: \: x = - 4 + 5 \\ \\ x = 9 \: \: \: \: \: \: \: \: x = 1](https://tex.z-dn.net/?f=+%7Bx%7D%5E%7B2%7D++-+10x+%2B+9+%3D+0+%5C%5C++%5C%5C++%7Bx%7D%5E%7B2%7D++-+10x+%3D++-+9+%5C%5C++%5C%5C+add+%5C%3A++%7B5%7D%5E%7B2%7D+on+%5C%3A+both+%5C%3A+sides+%5C%5C++%5C%5C++%7Bx%7D%5E%7B2%7D++-+2x%285%29+%2B++%7B5%7D%5E%7B2%7D++%3D++-+9+%2B++%7B5%7D%5E%7B2%7D++%5C%5C++%5C%5C++%7B%28x+-+5%29%7D%5E%7B2%7D++%3D++-+9+%2B+25+%5C%5C++%5C%5C+++%7B%28x+-+5%29%7D%5E%7B2%7D++%3D+16+%5C%5C++%5C%5C+x+-+5+%3D++%2B+and+%5C%3A++-++%5Csqrt%7B16%7D++%5C%5C++%5C%5C+x+-+5+%3D++%2B+and+%5C%3A++-+4+%5C%5C++%5C%5C+x+%3D+4+%2B+5+%5C%3A++%5C%3A++%5C%3A+x+%3D++-+4+%2B+5+%5C%5C++%5C%5C+x+%3D+9+%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A+x+%3D+1)
HOPE THIS HELPS U...
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Here is your answer :
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Given,
HOPE THIS HELPS U...
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