Advertisements
Advertisements
Question
In the following, determine whether the given values are solutions of the given equation or not:
x2 + x + 1 = 0, x = 0, x = 1
Solution
We have been given that,
x2 + x + 1 = 0, x = 0, x = 1
Now if x = 0 is a solution of the equation then it should satisfy the equation.
So, substituting x = 0 in the equation, we get
x2 + x + 1
= (0)2 + 0 + 1
= 1
Hence x = 0is not a solution of the given quadratic equation.
Also, if x = 1is a solution of the equation then it should satisfy the equation.
So, substituting x = 1 in the equation, we get
x2 + x + 1
= (1)2 + 1 + 1
= 3
Hence x = 1 is not a solution of the quadratic equation.
Therefore, from the above results we find out that both x = 0 and x = 1are not a solution of the given quadratic equation.
APPEARS IN
RELATED QUESTIONS
Check whether the following is the quadratic equation:
(x - 2)(x + 1) = (x - 1)(x + 3)
Solve x2 + 7x = 7 and give your answer correct to two decimal places
Find the quadratic equation, whose solution set is:
{5, −4,}
Which of the following are quadratic equation in x?
`x-6/x=3`
`x^2-(1+sqrt2)x+sqrt2=0`
Is x = –3 a solution of the quadratic equation 2x2 – 7x + 9 = 0?
Solve:
`sqrt(x/(x - 3)) + sqrt((x - 3)/x) = 5/2`
Solve:
`((3x + 1)/(x + 1)) + ((x + 1)/(3x + 1)) = 5/2`
Solve:
`1/(18 - x) - 1/(18 + x) = 1/24` and x > 0
Write the following quadratic equation in standard form ax2 + bx + c = 0 : x (x + 3) = 7