In each of the following, determine whether the given values are solutions of the given equation or not:
(i)x²-3x+2=0, x=2, x=-1
(ii)x²+x+1=0, x=0, x=1
(iii)x²-3√3x+6=0, x=√3, x=-2√3
Answers
SOLUTION :
(i)Given x² - 3x + 2 = 0 , x = 2 , x = -1
Equation is of the form p(x) = 0 , where
p(x) = x² - 3x + 2 ……………(1)
On putting x = 2 in eq 1,
p(2) = 2² - 3(2) + 2
= 4 - 6 + 2
= 6 - 6
p(2) = 0
So, x = 2 is a solution of the given equation.
Now On putting x = - 1 in eq 1,
p(x) = x² - 3x + 2
p(- 1) = (-1)² - 3(-1) + 2
= 1 + 3 + 2
= 6
p(- 1) ≠ 0
So, x = - 1 is not the solution of the given equation.
Hence, x = 2 is the solution of the given equation and x = - 1 is not the solution of the given equation
(ii) Given : x² + x + 1 = 0
Equation is of the form p(x) = 0 , where
p(x) = x² + x + 1 ……………(1)
On putting x = 0 in eq 1,
p(0) = 0² + 0 + 1
= 0 + 0 + 1
= 1
p(0) ≠ 0
So, x = 0 is not the solution of the given equation.
Now On putting x = 1 in eq 1,
p(x) = x² + x + 1
= 1² + 1 + 1
= 1 +1 + 1
= 3
p(1) ≠ 0
So, x = 1 is not the solution of the given equation.
Hence, x = 0 and x = 1 both are not the solution of the above equation
(iii) Given : x² - 3√3x + 6 = 0
Equation is of the form p(x) = 0 , where
p(x) = x² - 3√3x + 6 …………..(1)
On putting x = √3 in eq 1,
p(√3) = √3² - 3√3 × √ 3 + 6
= 3 – 9 + 6
= 9 - 9
= 0
p(√3) = 0
So, x = √3 is a solution of the given equation.
Now on putting x = -2√3 in eq 1,
p(-2√3) = (-2√3)² - 3√3 × -2√3 + 6
= 12 + 18 + 6
= 36
p(-2√3) ≠ 0
So, x = -2√3 is not the solution of the given equation.
Hence, x = √3 is a solution of the above equation and x = - 2√3 is not a solution of the above equation.
HOPE THIS ANSWER WILL HELP YOU…
1) Put x= -1 and put x=2 in the equations and if zero is comming then they are the solution of this equation.