Question

Find the value of k for which x = 1 is a root of the equation x^2 + kx + 3 = 0. Also, find the other root.

Answer 1

x^2+kx+3=0
now putting x=1
(1)^2)+k×1+3=0
1+k+3=0
k+4=0
k=-4
&
putting the value of k=-4 in eq.
x^2+(-4)x+3=0
x^2-4x+3=0
x^2-3x-x+3=0
x (x-3)-1 (x-3)=0
(x-1) (x-3)=0
x=1,x=3
I hope this helps you.
Ok
Be Brainy

Answer 2

Let the Given Quadratic Polynomial be : P(x)

⇒ P(x) = x² + kx + 3

Given that x = 1 is the Root of this Quadratic Equation

⇒ P(1) = 0

⇒ (1)² + k(1) + 3 = 0

⇒ k + 4 = 0

⇒ k = -4

⇒ The Given Quadratic Equation becomes : x² - 4x + 3 = 0

⇒ x² - 4x + 3 = 0

⇒ x² - 3x - x + 3 = 0

⇒ x(x - 3) - (x - 3) = 0

⇒ (x - 3)(x - 1) = 0

⇒ x = 3 or x = 1

So, The other root of Given Quadratic Equation is 3