Question

(a+b-2c)x²+(2a-b-c)x+(c+a-2b)=0 ..find the value of x​

Answer 1

$\huge{ \underline{ \bold{ \orange{ \mathfrak{question}}:}}}$

(a+b-2c)x²+(2a-b-c)x+(c+a-2b)=0 .. what is the value of x ?

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$\huge{{ \underline{ \bold{ \green{ \mathbb{SOLUTION}}:}}}} \:$

$(a + b - 2c) {x}^{2} + (2a - b - c)x + (c + a - 2b) = 0 \\ ⇛we \: know \: the \: formula \: for \: finding \\ out \:</strong><strong>x</strong><strong> </strong><strong>\:</strong><strong>v</strong><strong>a</strong><strong>l</strong><strong>u</strong><strong>e</strong><strong>s</strong><strong> </strong><strong>\:</strong><strong>o</strong><strong>r</strong><strong> </strong><strong>\:</strong><strong>e</strong><strong>l</strong><strong>s</strong><strong>e</strong><strong> </strong><strong>\:</strong><strong>we \: can \: factorise \: the \: middle \: term$

$\bold{ \overline{ \underline{ \underline{ \red{first \: method}}}:}}$

$⇛(a + b - 2c) {x}^{2} + ((a + b - 2c) + (c + a - 2b)x +( c + a - 2b) = 0 \\ ⇛(a + b - 2c) {x}^{2} + ((a + b - 2c)x + (c + a - 2b)x +( c + a - 2b) = 0 \\ ⇛(a + b - 2c)x(x + 1) + (c + a - 2b)(x + 1) = 0 \\ take \:( x + 1) \: common \: we \: get \: \\ ⇛(x + 1)((a + b - 2c)x + (c + a - 2b) = 0 \\ therefore \\ ⇛x = - 1 \: and \: \frac{ - (c + a - 2b)}{a + b - 2c} = \frac{2b - a - c}{a + b - 2c}$

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$\bold{ \overline{ \underline{ \underline{ \blue{second \: method}}}:}}$

$we \: can \: get \: the \: same \: solution \: \\ by \: applying \: quadratic \: formula \: \\ which \: is \: \\ x = \frac{ - b \: + | \sqrt{ {b}^{2} - 4ac } | }{2a} \\ and \: here \: also \: we \: get \: \\ x = - 1 \: and \: \frac{2b - c - a}{a + b - 2c}$

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$\underline{\overline{\underline\bold \red{final \: answer}}}$

$x = - 1 \: and \: \frac{2b - c - a}{a + b - 2c} \:$

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