Answer to Question #285504 in Algebra for bookaddict

Question

find the values of k for which the following equations have two equal roots.

(a)4x^2+4(k-2)x+k=0

(b)(k+2)x^2+4k=(4k+2x)

(c)x^2-2x+1=2k(k-2)

(d)(k+1)x^2+kx-2k=0

(e)4x^2-(k-2)x+9=0

Answer 1

(a)

$$4x^2+4(k-2)x+k=0$$

$$D=(4(k-2))^2-4(4)(k)=0$$

$$k^2-4k+4-k=0$$

$$k^2-5k+4=0$$

$$(k-1)(k-4)=0$$

$$k\in \{1, 4\}$$

(b)

$$(k+2)x^2+4k=4k+2x$$

$k\not=-2$

$$(k+2)x^2-2x=0$$

$$x_1=0, x_2=\dfrac{2}{k+2}\not=0, k\in \R, k\not=-2$$

There is no solution.

$$k\in \empty$$

(c)

$$x^2-2x+1=2k(k-2)$$

$$(x-1)^2=2k(k-2)$$

$$2k(k-2)=0$$

$$k_1=0, k_2=2$$

$$k\in \{0,2\}$$

(d)

$$(k+1)x^2+kx-2k=0$$

$k\not=-1$

$$D=(k)^2-4(k+1)(-2k)=0$$

$$k^2+8k+8=0$$

$$k^2+8k+16=8$$

$$(k+4)^2=8$$

$$k_1=-2\sqrt{2}, k_2=2\sqrt{2}$$

$$k\in \{-2\sqrt{2},2\sqrt{2}\}$$

(e)

$$4x^2-(k-2)x+9=0$$

$$D=(-(k-2))^2-4(4)(9)=0$$

$$(k-2)^2=144$$

$$k_1=-10, k_2=14$$

$$k\in \{-10,14\}$$