# If the equation x^2+ (y+1)x -1 =0 has equal roots what are the values of y

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### 3 Answers

**We'll try to calculate the value of y using Viete's relations:**

x1 + x2 = -b/a

x1*x2 = c/a

We'll identify the coefficients of the equation:

a = 1

b = y+1

c = -1

We also know, from enunciation, taht the roots are equal:x1=x2

x1 + x2 = 2x = - y - 1 (1)

x^2 = -1 (2)

We'll raise to square (1):

4x^2 = (y+1)^2

But x^2 = -1. We'll substitute x^2 by the value -1 and we'll expand the square from the right side:

-4 = y^2 + 2y + 1

We'll add 4 both sides and we'll use symmetric property:

y^2 + 2y + 1 + 4 = 0

y^2 + 2y + 5 = 0

We'll apply the quadratic formula:

y1 = [-2+sqrt(4-20)]/2

y1 = [-2 + sqrt(-16)]/2

y1 = (-2+4i)/2

**y1 = -1 + 2i**

**y2 = -1 - 2i**

As the equation x^2+ (y+1) x - 1 =0 has only equal roots we know that this is possible only if b^2 – 4ac=0 where b and c are a coefficients of x, the numeric term and x^2 resp.

From the equation b= y+1, a =1 and c =-1

=> (y + 1) ^2 + 4* 1*1 =0

=> Y^2 + 2y +1 + 4 =0

=> Y^2 + 2y +5 =0

The roots of y are [-2 + sqrt (4- 20)/2 and [-2 – sqrt (4-20)/2

or -1+2i and -1-2i

**So if y = -1+2i or -1-2i, x^2+ (y+1)x - 1 =0 has equal roots.**

x^2+(y+1)x-1 = 0 has equal roots.

To find the values of y.

Solution:

A quadratic equation ax^2+bx+c = 0 can have equal roots when the discriminant b^2-4ac = 0

Here a = 1, b = y+1 and c = -1.

So the discriminant of the given quadratic is (y+1)^2 - 4*1(-1).

There( y+1)^2 +4 = 0 . But (y+1)^2 +2^2 = 0 is not true as the minimum value of the discriminant is 4 when y = -1. So the discriminant is always positive for all real values of y.

Therefore the given expression x^2+(y+1)x -1 = 0 cannot have two equal roots for any real value of y.