Advertisements
Advertisements
प्रश्न
In each of the following determine whether the given values are solutions of the equation or not.
x2 + x + 1 = 0; x = 1, x = -1.
उत्तर
Given equation is
x2 + x + 1 = 0; x = 1, x = -1
Substitute x = 1 in L.H.S.
L.H.S. = (1)2 + (1) + 1
= 3 ≠ R.H.S. ≠ 0
Hence, x = 1 is not a solution of the given equation.
Now substitute x = -1 in L.H.S.
L.H.S. = (-1)2 + (-1) + 1
= 1 - 1 + 1
= 1 ≠ R.H.S. ≠ 0
Hence, x = -1 is not a solution of the given equation.
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equations by factorization:
`(x-a)/(x-b)+(x-b)/(x-a)=a/b+b/a`
Solve the following quadratic equations by factorization:
`7x + 3/x=35 3/5`
Find the two consecutive positive even integers whose product is 288.
The difference of two natural numbers is 5 and the difference of heir reciprocals is `5/14`Find the numbers
One of the roots of equation 5m2 + 2m + k = 0 is `(-7)/5` Complete the following activity to find the value of 'k'.
A two digit number is such that the product of its digit is 8. When 18 is subtracted from the number, the digits interchange its place. Find the numbers.
The hypotenuse of a grassy land in the shape of a right triangle is 1 m more than twice the shortest side. If the third side is 7m more than the shortest side, find the sides of the grassy land.
Solve the following equation by factorization
`4sqrt(3)x^2 + 5x - 2sqrt(3)` = 0
Sum of two natural numbers is 8 and the difference of their reciprocal is 2/15. Find the numbers.
Using quadratic formula find the value of x.
p2x2 + (p2 – q2)x – q2 = 0