Dear student, b+cx2-a+b+cx+a=0D=a+b+c2-4b+ca=a2+b2+c2-2ab+2bc-2ca=a-b-c2x=a+b+c±a-b-c22b+c=a+b+c±a-b-c2b+c∴x=ab+cor 1 Hope it would have clear your doubt Do get back to us in case of any further concerns

(b+c)x²-(a+b+c)x+a=0

Dear student,

(b + c) x^{2} - (a + b + c) x + a = 0 D = {(a + b + c)}^{2} - 4 (b + c) a = a^{2} + b^{2} + c^{2} - 2 a b + 2 b c - 2 c a = {(a - b - c)}^{2} x = \frac{(a + b + c) \pm \sqrt{{(a - b - c)}^{2}}}{2 (b + c)} = \frac{(a + b + c) \pm (a - b - c)}{2 (b + c)} ∴ x = \frac{a}{b + c} or 1

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(b+c)x2-(a+b+c)x+a=0

(b+c)x²-(a+b+c)x+a=0