Advertisements
Advertisements
प्रश्न
In each of the following determine whether the given values are solutions of the equation or not.
x2 + 6x + 5 = 0; x = -1, x = -5
Advertisements
उत्तर
Given equation is
x2 + 6x + 5 = 0; x = -1, x = -5
Substitute x = -1 in L.H.S.
L.H.S. = (-1)2 + 6 x (-1) + 5
= 1 - 6 + 5
= 6 - 6
= 0
Hence, x = -1 is a solution of the given equation.
Again put x = -5 in L.H.S.
L.H.S. = (-5)2 + 6 x (-5) + 5
= 25 - 30 + 5
= 30 - 30
= 0
Hence, x = -5 is also a solution of the given equation.
संबंधित प्रश्न
Solve for x :
`1/(x + 1) + 3/(5x + 1) = 5/(x + 4), x != -1, -1/5, -4`
Two number differ by 4 and their product is 192. Find the numbers?
Sum of the areas of two squares is 640 m2. If the difference of their perimeters is 64 m. Find the sides of the two squares.
Without solving the following quadratic equation Find the value of p for which the roots are equal
`px^2 - 4x + 3 = 0`
Solve the following quadratic equation by factorisation.
25m2 = 9
Solve the following quadratic equations by factorization: \[\frac{2}{x + 1} + \frac{3}{2(x - 2)} = \frac{23}{5x}; x \neq 0, - 1, 2\]
Solve the following quadratic equation using formula method only
x2 - 6x + 4 = 0
Solve the following equation by factorization
6p2+ 11p – 10 = 0
Find the values of x if p + 1 =0 and x2 + px – 6 = 0
If x = –2 is the common solution of quadratic equations ax2 + x – 3a = 0 and x2 + bx + b = 0, then find the value of a2b.
