Question

 If the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal, then prove that 2b = a + c.​

Answer 1

Step-by-step explanation:

If roots of a quadratic equation are equal, then discriminant of the quadratic equation is 0

$D= {b}^{2} −4ac=0$

$(b−c) {x}^{2} +(c−a)x+(a−b)=0$

$Comparing \: with$

$a {x}^{2} +bx+c=0$

$Here, a=(b−c), b=(c−a) and c=(a−b)$

$So, </strong></p><p></p><p><strong>[tex]So, </strong></p><p><strong>⇒ {(c−a)}^{2} −4(b−c)(a−b)=0</strong></p><p></p><p><strong>[tex]So, ⇒ {(c−a)}^{2} −4(b−c)(a−b)=0$

$⇒ {c}^{2} + {a}^{2} −2ac−4(ab− {b}^{2} −ac+bc)=0$

$⇒ {c}^{2} + {a}^{2} −2ac−4ab+4 {b}^{2} +4ac−4bc=0$

$⇒ {c}^{2} + {a}^{2} +2ac+4 {b}^{2} −4ab−4bc=0$

$⇒ {(c+a)}^{2} +4 {b}^{2} −4b(a+c)=0$

$⇒ {(c+a)}^{2} + {(2b)}^{2} −2(c+a)(2b)=0$

$⇒ {[(c+a)−(2b)]}^{2} =0$

$⇒c+a−2b=0$

$⇒2b=c+a$

Hope it may help you....

Answer 2

Answer:

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