Advertisements
Advertisements
Question
In each of the following determine whether the given values are solutions of the equation or not.
9x2 - 3x - 2 = 0; x = `-(1)/(3), x = (2)/(3)`
Advertisements
Solution
Given equation is
9x2 - 3x - 2 = 0; x = `-(1)/(3), x = (2)/(3)`
Substitute x = `-(1)/(3)` in the L.H.S.
L.H.S. = `9(-1/3)^2 - 3 xx (-1/3) -2`
= `9 xx (1)/(9) + 1 - 2`
= 2 - 2
= 0
= R.H.S.
Hence, x = `-(1)/(3)` is a solution of the equation.
Again put x = `(2)/(3)`
L.H.S. = `9(2/3)^2 -3(2/3)-2`
= `9 xx (4)/(9) - 2 -2`
= 4 - 4
= 0
= R.H.S.
Hence, x = `(2)/(3)` is a solution of the equation.
RELATED QUESTIONS
Solve for x :
`3/(x+1)+4/(x-1)=29/(4x-1);x!=1,-1,1/4`
John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
Solve the following quadratic equations by factorization:
(a + b)2x2 - 4abx - (a - b)2 = 0
The sum of the squares to two consecutive positive odd numbers is 514. Find the numbers.
Solve the following quadratic equations by factorization: \[\frac{5 + x}{5 - x} - \frac{5 - x}{5 + x} = 3\frac{3}{4}; x \neq 5, - 5\]
Solve the following equation: 4x2 - 13x - 12 = 0
Write the number of zeroes in the end of a number whose prime factorization is 22 × 53 × 32 × 17.
Find the values of x if p + 7 = 0, q – 12 = 0 and x2 + px + q = 0,
2x articles cost Rs. (5x + 54) and (x + 2) similar articles cost Rs. (10x – 4), find x.
Find the roots of the following quadratic equation by the factorisation method:
`21x^2 - 2x + 1/21 = 0`
