Question

4/ If the roots of the quadratic equation (x – a) (x – b) + (x – b)(x - c) + (x – C)(x – a) = 0 are equal. Then,
show that a = b = c.​

Answer 1

EXPLANATION.

Roots of the quadratic equation.

⇒ (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0.

As we know that,

Expand the equation, we get.

⇒ (x² - bx - ax + ab) + (x² - cx - bx + bc) + (x² - ax - cx + ac) = 0.

⇒ x² - bx - ax + ab + x² - cx - bx + bc + x² - ax - cx + ac = 0.

⇒ 3x² - 2ax - 2bx - 2cx + ab + bc + ac = 0.

⇒ 3x² - 2(a + b + c)x + (ab + bc + ac) = 0.

As we know that,

⇒ D = Discriminant  Or  b² - 4ac.

⇒ [-2(a + b + c)² - 4(3)(ab + bc + ac)] = 0.

⇒ [4(a + b + c)² - 4(3)(ab + bc + ac)] = 0.

⇒ [4(a² + b² + c² + 2ab + 2bc + 2ac) - 12(ab + bc + ac)] = 0.

⇒ [4(a² + b² + c² + 2ab + 2bc + 2ac) - 12ab - 12bc - 12ac ] = 0.

⇒ [4a² + 4b² + 4c² + 8ab + 8bc + 8ac - 12ab - 12bc - 12ac] = 0.

⇒ [4a² + 4b² + 4c² - 4ab - 4bc - 4ac] = 0.

⇒ 2[2a² + 2b² + 2c² - 2ab - 2bc - 2ac] = 0.

⇒ 2[(a - b)² + (b - c)² + (c - a)²] = 0.

⇒ a - b = 0.

⇒ a = b.

⇒ b - c = 0.

⇒ b = c.

⇒ c - a = 0.

⇒ c = a.

Hence proved = a = b = c.

                                                                                                                           

MORE INFORMATION.

Quadratic expression.

A polynomial of degree two of the form ax² + bx + c (a ≠ 0) is called a quadratic expression in x.

The quadratic expression.

ax² + bx + c = 0 (a ≠ 0) has two roots, given by.

⇒ α = - b + √D/2a.

⇒ β = - b - √D/2a.

⇒ D = Discriminant  Or  b² - 4ac.

Answer 2

Given :

$\bf{\green{(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0}}$

To find :

$\bf{\blue{Show \: that \: a = b = c.}}$

Solution :

Let us discriminate each side of the problem .

$\bf{\red{Equation =(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0}}$

First we should rearrange the equation

$\sf⇒x² - (a + b)x + ab + x² - (b + c)x + bc + x² - (c + a)x + ca$

$\sf ⇒4{a² + b² + c² + 2(ab + bc + ca)} -12(ab + bc + ca) = 0$

$\sf⇒3x² - 2(a + b + c)x + (ab + bc + ca)$

$\sf {2(a + b + c)}² - 4(ab + bc + ca).3 = 0$

$\sf ⇒(a - b)² + (b - c)² + (c - a)² = 0$

$\textsf {Hence , We got that , \red {a = b = c}}$