Question

find the nature of root of x2-x-1 =0​

Answer 1

Answer:

1.618  and  -0.618

Step-by-step explanation:

Address the input parameters and values

Quadratic Equation : x2 - x - 1 = 0

Substitute a , b, and c values in below formula

x =(- b ± √b² - 4ac )  /  2a

=(-(-1) ± √(-12 - 4 (1) (-1)) )   /  2(1)

=1± √1 - 4 (1)(-1)    /  2

=1± √1 + 4   /  2

=1± √5   /  2

Solve the above expression for minus & plus values

x1 =1 + √5   /  2

=1 + 2.2361   /  2

x2 =1 - √5   /  2

=1 - 2.2361   /  2

By simplifying the above expressions for x1 & x2

x1 =3.2361  / 2

x1 = 1.618

x2 =-1.2361 /2

x2 = -0.618

Therefore,  

x1 = 1.618 & x2 = -0.618 are the roots for equation x2 - x - 1 = 0.

HOPE IT HELPS YOU...

Answer 2

Answer:

Nature of root of $x^{2} -x -1 = 0$ is real .

Step-by-step explanation:

Explanation :

Given , $x^{2} -x -1 = 0$

Now we compare the given equation with quadratic equation  which is $ax^{2} +bx + c= 0$

Step 1:

So , on comparing  here we can see that

a = 1 , b = -1 and c = -1

Now from the formula of D = $b^{2} - 4ac$ we get ,

D = $(-1)^{2} - 4 .1 .(-1)$ = 5

Final answer :

Hence ,  roots of the given equation are  real .

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