Advertisements
Advertisements
प्रश्न
In each of the following, determine whether the given numbers are roots of the given equations or not; 6x2 – x – 2 = 0; `(-1)/(2), (2)/(3)`
Advertisements
उत्तर
6x2 – x – 2 = 0; `(-1)/(2), (2)/(3)`
If x = `(-1)/(2)`, then
= `6(-1/2)^2 - (-1/2) - 2`
= `6 xx (1)/(4) + (1)/(2) - 2`
= `(3)/(2) + (1)/(2) - 2`
= `(4)/(2) - 2` = 0
∴ x = `(-1)/(2)` is its root
If x = `(2)/(3)`, then
= `6 xx (4)/(9) - (2)/(3) - 2`
= `(8)/(3) - (2)/(3) - 2`
= `(6)/(3) - 2` = 0
∴ x = `(2)/(3)` is also its root.
Hence `(-1)/(2), (2)/(3)` are both its root.
APPEARS IN
संबंधित प्रश्न
Without solving, examine the nature of roots of the equation 2x2 – 7x + 3 = 0
Solve the following quadratic equation using formula method only
`2x^2 - 2 . sqrt 6x + 3 = 0`
Solve the following quadratic equation using formula method only
`3"x"^2 + 2 sqrt 5x - 5 = 0 `
Solve the following quadratic equation using formula method only
16x2 - 24x = 1
Find the value of k for which the roots of the equation 3x2 - 10x + k = 0 are reciprocal of each other.
In the quadratic equation kx2 − 6x − 1 = 0, determine the values of k for which the equation does not have any real root.
Determine whether the given quadratic equations have equal roots and if so, find the roots:
3x2 - 6x + 5 = 0
If a is a root of the equation x2 – (a + b)x + k = 0, find the value of k.
If b2 – 4ac > 0 and b2 – 4ac < 0, then write the nature of roots of the quadratic equation for each given case
The roots of the quadratic equation `1/("a" + "b" + "x") = 1/"a" + 1/"b" + 1/"x"`, a + b ≠ 0 is:
