Advertisements
Advertisements
प्रश्न
In each of the following find the values of k of which the given value is a solution of the given equation:
x2 + 3ax + k = 0; x = a.
उत्तर
x2 + 3ax + k = 0; x = a.
Putting x = -a in L.H.S. of equation
L.H.S = (-a)2 + 3a x (-a) + k = 0
⇒ a2 - 3a2 + k = 0
⇒ -2a2 + k = 0
Hence, k = 2a2.
APPEARS IN
संबंधित प्रश्न
Check whether the following is the quadratic equation:
x2 + 3x + 1 = (x - 2)2
Find the value of ‘p’, if the following quadratic equation have equal roots:
4x2 – (p – 2)x + 1 = 0
Solve the following equation using the formula:
2x2 + 7x + 5 = 0
Solve the following equation using the formula:
3x2 + 2x – 1 = 0
`48x^2-13x1=0`
Solve equation using factorisation method:
`9/2 x = 5 + x^2`
Solve the following equation by using formula :
`2x^2 + sqrt(5) - 5` = 0
Solve:
`2((2x - 1)/(x + 3)) - 3((x + 3)/(2x - 1)) = 5; x ≠ -3, (1)/(2)`
If x2 – 2x – 3 = 0; values of x correct to two significant figures are ______.
If x2 – 4x = 5, the value of x is ______.
