Question

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

Answer 1

Hint: 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. $D = {B^2} - 4AC = 0$, 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, $\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
$\left( a \right)$ If the roots of the equation, $\left( {b - c} \right){x^2} + \left( {c - a} \right)x + \left( {a - b} \right) = 0$ 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. $D = {B^2} - 4AC = 0$.
Where, A = b – c, B = c – a, and C = a – b.
Now substitute the values we have,
$ \Rightarrow D = {\left( {c - a} \right)^2} - 4\left( {b - c} \right)\left( {a - b} \right) = 0$
Now simplify using the property that ${\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy$ so we have,
$ \Rightarrow {c^2} + {a^2} - 2ac - 4ab + 4{b^2} + 4ac - 4bc = 0$
$ \Rightarrow {c^2} + {a^2} + 2ac - 4ab + 4{b^2} - 4bc = 0$
$ \Rightarrow {\left( {a + c} \right)^2} + 4{b^2} - 4b\left( {a + c} \right) = 0$, $\left[ {\because {c^2} + {a^2} + 2ac = {{\left( {a + c} \right)}^2}} \right]$
So it is the formula of ${\left( {x - y} \right)^2}$, where x = a + c and y = 2b
$ \Rightarrow {\left[ {\left( {a + c} \right) - 2b} \right]^2} = 0$
$ \Rightarrow a + c - 2b = 0$
$ \Rightarrow 2b = a + c$
Hence a, b, and c are in Arithmetic progression.
Hence proved.
$\left( b \right)$ If $a\left( {b - c} \right){x^2} + b\left( {c - a} \right)x + c\left( {a - b} \right) = 0$ 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. $D = {B^2} - 4AC = 0$.
Where, A = a(b – c), B = b(c – a), and C = c(a – b).
Now substitute the values we have,
$ \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 ${\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy$ so we have,
\[ \Rightarrow {b^2}\left( {{c^2} + {a^2} - 2ac} \right) - 4ac\left( {ab - {b^2} - ac + bc} \right) = 0\]
\[ \Rightarrow {b^2}{c^2} + {b^2}{a^2} - 2a{b^2}c - 4{a^2}bc + 4a{b^2}c + 4{a^2}{c^2} - 4ab{c^2} = 0\]
\[ \Rightarrow {b^2}{c^2} + {b^2}{a^2} + 2a{b^2}c - 4{a^2}bc + 4{a^2}{c^2} - 4ab{c^2} = 0\]
\[ \Rightarrow {\left( {ab + bc} \right)^2} + 4{a^2}{c^2} - 4ac\left( {ab + bc} \right) = 0\], $\left[ {\because {b^2}{c^2} + {b^2}{a^2} + 2a{b^2}c = {{\left( {ab + bc} \right)}^2}} \right]$
So it is the formula of ${\left( {x - y} \right)^2}$, where x = ab + bc and y = 2ac
\[ \Rightarrow {\left[ {\left( {ab + bc} \right) - 2ac} \right]^2} = 0\]
\[ \Rightarrow \left( {ab + bc} \right) - 2ac = 0\]
\[ \Rightarrow 2ac = ab + bc\]
Now divide by abc throughout we have,
\[ \Rightarrow \dfrac{{2ac}}{{abc}} = \dfrac{{ab}}{{abc}} + \dfrac{{bc}}{{abc}}\]
\[ \Rightarrow \dfrac{2}{b} = \dfrac{1}{c} + \dfrac{1}{a}\]
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 ${x^2}$, and the product of the roots is the ratio of the constant term to the coefficient of ${x^2}$, as the roots are equal so we can easily eliminate the variable, for example in $\left( {b - c} \right){x^2} + \left( {c - a} \right)x + \left( {a - b} \right) = 0$ let the roots be p and p.
So, the sum of the roots, p + p = $\dfrac{{ - \left( {c - a} \right)}}{{b - c}}$, therefore, 2p = $\dfrac{{ - \left( {c - a} \right)}}{{b - c}}$, and the product of the roots, ${p^2} = \dfrac{{a - b}}{{b - c}}$. So we can easily eliminate the variable p, by substituting $p = \dfrac{{ - \left( {c - a} \right)}}{{2\left( {b - c} \right)}}$ in second equation so we have, $\left[ {{{\left( {\dfrac{{ - \left( {c - a} \right)}}{{2\left( {b - c} \right)}}} \right)}^2} = \dfrac{{a - b}}{{b - c}}} \right]$, $ \Rightarrow \dfrac{{{{\left( {c - a} \right)}^2}}}{{4\left( {b - c} \right)}} = a - b$, rest of the solution remains the same.