Question

x²+4x-5=0,x=1,-1 Determine whether the values given against each of the quadratic equation are the roots of the equation.

Answer 1

Solution:-

given by quadratic equation:-
${x}^{2} + 4x - 5 = 0 \\ {x}^{2} + 5x - x - 5 = 0 \\ x(x + 5) - 1(x + 5) = 0 \\ (x + 5)(x - 1) = 0 \\ x = - 5 \\ x = 1$
here , x = (-5 ,1)

●according to question given by zeroes( -1) does not exits.

●here , x= ( 1) is the root of given quadratic equation is (x²+4x-5=0)

☆i hope its help☆

Answer 2

METHOD TO CHECK WHETHER THE GIVEN VALUE IS A ZERO OF A POLYNOMIAL OR NOT:
If a polynomial in one variable (x) is given to us and a value of variable (x = c) is also given then to check that a value of x is a zero of given polynomial or not, we use the following steps.
•Put x = c in given polynomial and find the value of p(c).
• If p(c)= 0 , then x = c will be a zero of given polynomial and if p(c) ≠ 0, then x= c will not be a zero of a given polynomial.

SOLUTION :
Let p(x) = x²+4x-5…...........(1)

On Putting x = 1 in eq 1
x²+4x-5
(1)² + 4 × 1 - 5 = 1 +4 -5 = 5 -5 =0
Here, x=1 is the root of p(x).

On Putting x = -1 in eq 1.
x²+4x-5
(-1)² + 4 (-1) -5 = 1 - 4 -5 = -3 -5 = - 8
Here, x(-1) ≠ 0

Hence,x= 1 is a root of the given equation & x = -1 is not a root of the Given equation.

HOPE THIS WILL HELP YOU..