Question

★ Find the value of k. So that, the Quadratic equation have real roots

=> y² + k² = 2(k+1)y

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Answer 1

The following quadrilateral is given.

$y^2+k^2=2(k+1)y\\y^2-2(k+1)y+k^2=0$

We can interpret from it the follows.

$a=1\\b=-2(k+1)\\c=k^2$

Now they said that it has 2 real roots.

$D={b^2-4ac} >0$

${[-2(k+1)]^2-4k^2} >0$

$[-2(k+1)]^2>4k^2$

Square roots on both sides

$-2(k+1)>2k$

$-k-1>k\\2k>-1\\k>-\frac{1}{2}$

∴ $k>-\frac{1}{2}$

Answer 2

Step-by-step explanation:

Consider the given equation.

2x

2

+5x+k=0 .......(1)

Since, the equation has no real roots.

b

2

−4ac<0

Therefore,

5

2

−4×2×k<0

25−8k<0

25<8k

k>

8

25

Hence, this is the answer.