Question

please answer this quadratic equation
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Answer 1

Answer:

2a=b+c by using b2 -4ac

Step-by-step explanation:

${b }^{2} - 4ac = 0$

${b }^{2} - 4ac = 0$

a=a-b

b= b- c

c= c-a

by using the formula, replace a,b and c

${(b - c)}^{2} - 4(a - b)(c - a) = 0 \\ {b}^{2} + {c}^{2} - 2bc - 4(ac - {a}^{2} - bc + ba) = 0 \\ {b}^{2} + {c}^{2} - 2bc - 4c + 4 {a}^{2} + 4bc - 4ba = 0 \\ 4 {a}^{2} + {b}^{2} + {c}^{2} + 2bc - 4ac - 4ba = 0 \\ {2}^{2} {a }^{2} + {b}^{2} + {c}^{2} + 2bc - 2 \times 2ac - 2 \times 2ba = 0 \\ {2a}^{2} + { (- b)}^{2} + {( - c)}^{2} + 2( - b)( - c) + 2 \times 2a( - c) + 2 \times 2a( - b) = 0 \\ {(2a - b - c)}^{2} = 0 \\ taking \: square \: root \: on \: both \: sides \\ 2a - b - c = 0 \\ 2a = b + c \\ hence \: proved</p><p>[tex]{(b - c)}^{2} - 4(a - b)(c - a) = 0 \\ {b}^{2} + {c}^{2} - 2bc - 4(ac - {a}^{2} - bc + ba) = 0 \\ {b}^{2} + {c}^{2} - 2bc - 4c + 4 {a}^{2} + 4bc - 4ba = 0 \\ 4 {a}^{2} + {b}^{2} + {c}^{2} + 2bc - 4ac - 4ba = 0 \\ {2}^{2} {a }^{2} + {b}^{2} + {c}^{2} + 2bc - 2 \times 2ac - 2 \times 2ba = 0 \\ {2a}^{2} + { (- b)}^{2} + {( - c)}^{2} + 2( - b)( - c) + 2 \times 2a( - c) + 2 \times 2a( - b) = 0 \\ {(2a - b - c)}^{2} = 0 \\ taking \: square \: root \: on \: both \: sides \\ 2a - b - c = 0 \\ 2a = b + c \\ hence \: proved$

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