Question

if roots of the equation (a-b)x²+(b-c)x +(c-a) =0 are equal then b+c=​

Answer 1

EXPLANATION.

Roots of the equation,

f(x) = ( a - b)² + ( b - c)x + ( c - a ) = 0.

To Find the value of b + c.

As we know that,

D = b² - 4ac.

⇒  (b - c)² - 4(a - b)(c - a) = 0.

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

⇒ b² + c² - 2bc - 4ac + 4a² + 4bc - 4ab = 0.

⇒ b² + c² + 2bc + 2a² - 4ac - 4ab = 0.

⇒ (2a - b - c )² = 0.

⇒ 2a = b + c.

Value of b + c = 2a.

                                                 

MORE INFORMATION.

Quadratic equation.

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

The Quadratic Equation,

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

⇒ α = -b + √b² - 4ac/2a

⇒ β = -b - √b² - 4ac/2a.

Sum of Roots.

The Quadratic Equation whose roots are α,β.

Sum of zeroes of quadratic equation.

⇒ α + β = -b/a.

Products of zeroes of quadratic equation.

⇒ αβ = c/a.

Quadratic equation ⇒ x² - ( α + β)x + αβ.

Answer 2

$\Huge\underline{\overline{\mid{\green{Answer}}\mid}}$

Hi,

We know that,

If the quadratic equation ax²+bx+c=0

whose roots are equal then it's

deteminant is equal to zero.

(a-b)x²+(b-c)x+(c-a)=0

Deteminant =0

(b-c)² -4(a-b)(c-a)==0

b²+c²-2bc-4ac+4a²+4bc-4ab=0

b²+c²+4a²+4bc-4ac-4ab=0

b²+c²+(-2a)²+2bc+2c(-2a)+2(-2a)b=0

(b+c-2a)²=0

b+c-2a=0

Therefore,

b+c=2a

Hence proved.

$\Huge \boxed{ \colorbox{gold}{hope \: this \: helps}}$

:)