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Question: \(\left( a \right)\) If the roots of the equation, \(\left( {b - c} \right){x^2} + \left( {c - a} \r...

(a)\left( a \right) If the roots of the equation, (b−c)x2+(c−a)x+(a−b)=0\left( {b - c} \right){x^2} + \left( {c - a} \right)x + \left( {a - b} \right) = 0 be equal, then prove that a, b, c are in arithmetic progression.
(b)\left( b \right) If a(b−c)x2+b(c−a)x+c(a−b)=0a\left( {b - c} \right){x^2} + b\left( {c - a} \right)x + c\left( {a - b} \right) = 0 has equal roots, prove that a, b, c are in harmonic progression.

Explanation

Solution

In this particular question use the concept that if a quadratic equation has equal roots then the value of discriminant D must be zero i.e. D=B2−4AC=0D = {B^2} - 4AC = 0, and use the concept that if a, b, and C are in A.P then it must satisfies the condition, 2b = a + c and if a, b, and C are in H.P then it must satisfies the condition, 2b=1a+1c\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}, so use these concepts to reach the solution of the question.

Complete step-by-step answer :
(a)\left( a \right) If the roots of the equation, (b−c)x2+(c−a)x+(a−b)=0\left( {b - c} \right){x^2} + \left( {c - a} \right)x + \left( {a - b} \right) = 0 be equal, then prove that a, b, c are in arithmetic progression.
Proof –
As roots are equal then the value of discriminant D must be zero i.e. D=B2−4AC=0D = {B^2} - 4AC = 0.
Where, A = b – c, B = c – a, and C = a – b.
Now substitute the values we have,
⇒D=(c−a)2−4(b−c)(a−b)=0\Rightarrow D = {\left( {c - a} \right)^2} - 4\left( {b - c} \right)\left( {a - b} \right) = 0
Now simplify using the property that (x−y)2=x2+y2−2xy{\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy so we have,
⇒c2+a2−2ac−4ab+4b2+4ac−4bc=0\Rightarrow {c^2} + {a^2} - 2ac - 4ab + 4{b^2} + 4ac - 4bc = 0
⇒c2+a2+2ac−4ab+4b2−4bc=0\Rightarrow {c^2} + {a^2} + 2ac - 4ab + 4{b^2} - 4bc = 0
⇒(a+c)2+4b2−4b(a+c)=0\Rightarrow {\left( {a + c} \right)^2} + 4{b^2} - 4b\left( {a + c} \right) = 0, [∵c2+a2+2ac=(a+c)2]\left[ {\because {c^2} + {a^2} + 2ac = {{\left( {a + c} \right)}^2}} \right]
So it is the formula of (x−y)2{\left( {x - y} \right)^2}, where x = a + c and y = 2b
⇒[(a+c)−2b]2=0\Rightarrow {\left[ {\left( {a + c} \right) - 2b} \right]^2} = 0
⇒a+c−2b=0\Rightarrow a + c - 2b = 0
⇒2b=a+c\Rightarrow 2b = a + c
Hence a, b, and c are in Arithmetic progression.
Hence proved.
(b)\left( b \right) If a(b−c)x2+b(c−a)x+c(a−b)=0a\left( {b - c} \right){x^2} + b\left( {c - a} \right)x + c\left( {a - b} \right) = 0 has equal roots, prove that a, b, c are in harmonic progression.
Proof –
As roots are equal then the value of discriminant D must be zero i.e. D=B2−4AC=0D = {B^2} - 4AC = 0.
Where, A = a(b – c), B = b(c – a), and C = c(a – b).
Now substitute the values we have,
⇒D=b2(c−a)2−4ac(b−c)(a−b)=0\Rightarrow D = {b^2}{\left( {c - a} \right)^2} - 4ac\left( {b - c} \right)\left( {a - b} \right) = 0
Now simplify using the property that (x−y)2=x2+y2−2xy{\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy so we have,
⇒b2(c2+a2−2ac)−4ac(ab−b2−ac+bc)=0\Rightarrow {b^2}\left( {{c^2} + {a^2} - 2ac} \right) - 4ac\left( {ab - {b^2} - ac + bc} \right) = 0
⇒b2c2+b2a2−2ab2c−4a2bc+4ab2c+4a2c2−4abc2=0\Rightarrow {b^2}{c^2} + {b^2}{a^2} - 2a{b^2}c - 4{a^2}bc + 4a{b^2}c + 4{a^2}{c^2} - 4ab{c^2} = 0
⇒b2c2+b2a2+2ab2c−4a2bc+4a2c2−4abc2=0\Rightarrow {b^2}{c^2} + {b^2}{a^2} + 2a{b^2}c - 4{a^2}bc + 4{a^2}{c^2} - 4ab{c^2} = 0
⇒(ab+bc)2+4a2c2−4ac(ab+bc)=0\Rightarrow {\left( {ab + bc} \right)^2} + 4{a^2}{c^2} - 4ac\left( {ab + bc} \right) = 0, [∵b2c2+b2a2+2ab2c=(ab+bc)2]\left[ {\because {b^2}{c^2} + {b^2}{a^2} + 2a{b^2}c = {{\left( {ab + bc} \right)}^2}} \right]
So it is the formula of (x−y)2{\left( {x - y} \right)^2}, where x = ab + bc and y = 2ac
⇒[(ab+bc)−2ac]2=0\Rightarrow {\left[ {\left( {ab + bc} \right) - 2ac} \right]^2} = 0
⇒(ab+bc)−2ac=0\Rightarrow \left( {ab + bc} \right) - 2ac = 0
⇒2ac=ab+bc\Rightarrow 2ac = ab + bc
Now divide by abc throughout we have,
⇒2acabc=ababc+bcabc\Rightarrow \dfrac{{2ac}}{{abc}} = \dfrac{{ab}}{{abc}} + \dfrac{{bc}}{{abc}}
⇒2b=1c+1a\Rightarrow \dfrac{2}{b} = \dfrac{1}{c} + \dfrac{1}{a}
Hence a, b, and c are in Harmonic progression.
Hence proved.

Note : We can also prove the conditions using the property that in a quadratic equation the sum of the roots is the ratio of the negative times the coefficient of x to the coefficient of x2{x^2}, and the product of the roots is the ratio of the constant term to the coefficient of x2{x^2}, as the roots are equal so we can easily eliminate the variable, for example in (b−c)x2+(c−a)x+(a−b)=0\left( {b - c} \right){x^2} + \left( {c - a} \right)x + \left( {a - b} \right) = 0 let the roots be p and p.
So, the sum of the roots, p + p = −(c−a)b−c\dfrac{{ - \left( {c - a} \right)}}{{b - c}}, therefore, 2p = −(c−a)b−c\dfrac{{ - \left( {c - a} \right)}}{{b - c}}, and the product of the roots, p2=a−bb−c{p^2} = \dfrac{{a - b}}{{b - c}}. So we can easily eliminate the variable p, by substituting p=−(c−a)2(b−c)p = \dfrac{{ - \left( {c - a} \right)}}{{2\left( {b - c} \right)}} in second equation so we have, [(−(c−a)2(b−c))2=a−bb−c]\left[ {{{\left( {\dfrac{{ - \left( {c - a} \right)}}{{2\left( {b - c} \right)}}} \right)}^2} = \dfrac{{a - b}}{{b - c}}} \right], ⇒(c−a)24(b−c)=a−b \Rightarrow \dfrac{{{{\left( {c - a} \right)}^2}}}{{4\left( {b - c} \right)}} = a - b, rest of the solution remains the same.