Advertisements
Advertisements
प्रश्न
If x = −3 and x = `2/3` are solutions of quadratic equation mx2 + 7x + n = 0, find the values of m and n.
उत्तर
Step 1: Sum and product of roots
Sum of roots = `-b/a, "Product of roots" = c/a`
- Roots: x1 = −3, x2 = `2/3`
- Coefficients: a = m, b = 7, c = n
Step 2: Calculate sum and product of roots
`x_1 + x_2 = -3 + 2/3 = -9/3 + 2/3 = -7/3`
Using Sum of roots = `−b/a`
`-7/3 = -7/m`
Equating the numerators, m = 3
Product of roots: `x_1xxx_2 = (-3)xx 2/3 = -2`
Using Product of roots = `c/a`
`-2 = n/m`
Substitute m = 3
`-2 = n/3 => n=-6`
m = 3, n = −6
APPEARS IN
संबंधित प्रश्न
Check whether the following is quadratic equation or not.
`x+1/x=x^2`, x ≠ 0
In the following, find the value of k for which the given value is a solution of the given equation:
7x2 + kx - 3 = 0, `x=2/3`
If x = 2/3 and x = −3 are the roots of the equation ax2 + 7x + b = 0, find the values of aand b.
Solve the following equation and give your answer correct to 3 significant figures: 5x² – 3x – 4 = 0
Solve `(2x)/(x - 3) + 1/(2x + 3) + (3x + 9)/((x - 3)(2x +3)) = 0; x != 3, x != - 3/2`
If quadratic equation x2 − (m + 1) x + 6 = 0 has one root as x = 3; find the value of m and the root of the equation.
Solve:
`x/a - (a + b)/x = (b(a + b))/(ax)`
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)`
Find the values of a and b from the quadratic equation 2x2 – 5x + 7 = 0.
