Advertisements
Advertisements
प्रश्न
In the following, find the value of k for which the given value is a solution of the given equation:
`kx^2+sqrt2x-4=0`, `x=sqrt2`
उत्तर
We are given here that,
`kx^2+sqrt2x-4=0`, `x=sqrt2`
Now, as we know that `x=sqrt2` is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting `x=sqrt2` in the above equation gives us,
`kx^2+sqrt2x-4=0`
`k(sqrt2)^2+sqrt2(sqrt2)-4=0`
2k + 2 - 4 = 0
2k - 2 = 0
2k = 2
`k=2/2`
k = 1
Hence, the value of k = 1.
APPEARS IN
संबंधित प्रश्न
Check whether the following is quadratic equation or not.
(x + 2)3 = x3 − 4
In the following, find the value of k for which the given value is a solution of the given equation:
x2 - x(a + b) + k = 0, x = a
Solve `(1 + 1/(x + 1))(1-1/(x - 1)) = 7/8`
Without solving the quadratic equation 6x2 – x – 2=0, find whether x = 2/3 is a solution of this equation or not.
Find the quadratic equation, whose solution set is:
{5, −4,}
`9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0`
The perimeter of a rectangular field is 82m and its area is 400m2, find the dimension of the rectangular field.
Find which of the following equations are quadratic:
(x – 1)2 + (x + 2)2 + 3(x + 1) = 0
Solve:
`((3x + 1)/(x + 1)) + ((x + 1)/(3x + 1)) = 5/2`
If x2 – 3x + 2 = 0, values of x correct to one decimal place are ______.
