What are the solutions to the equation (y+k)^2-2y^2-2(k+1)y-10=0 ?

The solutions to the equation (y+k)^2-2y^2-2(k+1)y-10=0 are y=-1-sqrt(k^2-9),y=-1+sqrt(k^2-9)

Find the zeros of (y+k)^2-2y^2-2(k+1)y-10=0

The zeros of (y+k)^2-2y^2-2(k+1)y-10=0 are y=-1-sqrt(k^2-9),y=-1+sqrt(k^2-9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y + k)^{2} - 2 y^{2} - 2 (k + 1) y - 10 = 0

y = - 1 - k^{2} - 9, y = - 1 + k^{2} - 9

Solution steps

(y + k)^{2} - 2 y^{2} - 2 (k + 1) y - 10 = 0

Expand

(y + k)^{2} - 2 y^{2} - 2 (k + 1) y - 10 : - y^{2} - 2 y + k^{2} - 10

- y^{2} - 2 y + k^{2} - 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) ( k ^{2} - 10 )}{2 ( - 1 )}

Simplify

(- 2)^{2} - 4 (- 1) (k^{2} - 10) : 2 k^{2} - 9

y_{1, 2} = \frac{- ( - 2 ) \pm 2 k ^{2} - 9}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- ( - 2 ) + 2 k ^{2} - 9}{2 ( - 1 )}, y_{2} = \frac{- ( - 2 ) - 2 k ^{2} - 9}{2 ( - 1 )}

y = \frac{- ( - 2 ) + 2 k ^{2} - 9}{2 ( - 1 )} : - 1 - k^{2} - 9

y = \frac{- ( - 2 ) - 2 k ^{2} - 9}{2 ( - 1 )} : - 1 + k^{2} - 9

The solutions to the quadratic equation are:

y = - 1 - k^{2} - 9, y = - 1 + k^{2} - 9

n = 60 h^{2}

x, (x^{2} + 4 x + 20) \cdot y = 23 sin (b) - 15 cos (t)

43 \cdot x^{2} - 2 x - 23 = 0

y, (t) [17 (2 - y) - 5 (y + 12)] (1 - 7 y) = 8

3 \cdot x^{2} + 10 x + 73 = 0

What are the solutions to the equation (y+k)^2-2y^2-2(k+1)y-10=0 ?
The solutions to the equation (y+k)^2-2y^2-2(k+1)y-10=0 are y=-1-sqrt(k^2-9),y=-1+sqrt(k^2-9)
Find the zeros of (y+k)^2-2y^2-2(k+1)y-10=0
The zeros of (y+k)^2-2y^2-2(k+1)y-10=0 are y=-1-sqrt(k^2-9),y=-1+sqrt(k^2-9)