Advertisements
Advertisements
प्रश्न
Find the values of x if p + 1 =0 and x2 + px – 6 = 0
Advertisements
उत्तर
p + 1 = 0, then p = – 1
Substituting the value of p in the given quadratic equation
x2 + ( – 1)x – 6 = 0
⇒ x2 – x – 6 = 0
⇒ x2 – 3x + 2x – 6 = 0
⇒ x(x – 3) + 2(x – 3) = 0
⇒ (x – 3) (x + 2) = 0
Either x – 3 = 0,
then x = 3
or
x + 2 = 0,
then x = – 2
Hence x = 3, -2.
APPEARS IN
संबंधित प्रश्न
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.
Solve the following quadratic equations by factorization:
`x^2 – (a + b) x + ab = 0`
In the following determine the set of values of k for which the given quadratic equation has real roots: \[2 x^2 + x + k = 0\]
Find the values of k for which the roots are real and equal in each of the following equation:\[px(x - 3) + 9 = 0\]
If 2 is a root of the equation x2 + ax + 12 = 0 and the quadratic equation x2 + ax + q = 0 has equal roots, then q =
The hypotenuse of a right-angled triangle is 17cm. If the smaller side is multiplied by 5 and the larger side is doubled, the new hypotenuse will be 50 cm. Find the length of each side of the triangle.
Solve the following equation by factorisation :
x2 + 6x – 16 = 0
Divide 16 into two parts such that the twice the square of the larger part exceeds the square of the smaller part by 164.
Find the roots of the following quadratic equation by the factorisation method:
`2/5x^2 - x - 3/5 = 0`
