Question

for what value of k the roots of quadratic equation y^2-2(k+1)y+k^2​

Answer 1

EXPLANATION.

Value of k of the roots of equation,

⇒ f(y) = y² - 2(k + 1)y + k².

As we know that,

⇒ D = b² - 4ac = 0.

⇒ [-2( k + 1)²] - 4(1)(k²) = 0.

⇒ [4(k² + 1 + 2k] - 4k² = 0.

⇒ 4k² + 4 + 8k - 4k² = 0.

⇒ 4 + 8k = 0.

⇒ 8k = -4.

⇒ k = -1/2.

Value of k = -1/2.

                                                                                         

MORE INFORMATION.

Maximum and minimum value of Quadratic Equation.

In a quadratic expression,

(1) = If a > 0, quadratic expression has least value at x = -b/2a. This least value is given by ⇒ 4ac - b²/4a = -D/2a.

(2) = If a < 0, quadratic expression has greatest value at x = -b/2a. This greatest value is given by ⇒ 4ac - b²/4a = -D/4a.

Answer 2

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Here is solution for your question bro...

Y2 + K2 - 2 ( K - 1 ) Y = 0

Y2 - 2( K-1 ) Y + K2 = 0

A = 1, B = -2( K - 1) , C = K2

GIVEN ROOTS ARE EQUAL, THEN B2 - 4AC = 0

{ - 2 ( K - 1 ) } SQUARE - 4 * 1 * K2 = 0

4 ( K2 - 2K + 1) - 4K2 = 0

4 { K2 - 2K + 1 - K2} = 0

4 { 1 - 2K } = 0

I - 2K = 0/4 = 0

1 = 2K

1/2 = K

K=1/2

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