Advertisements
Advertisements
प्रश्न
`3x^2-243=0`
उत्तर
Given:
`3x^2-243=0`
`⇒3(x^2-81)=0`
`⇒ (x)^2-(9)^2=0 `
`⇒ (x+9)(x-9)=0`
`⇒ x+9=0 or x-9=0`
⇒ x=-9 or x=9
Hence, -9 and 9 are the roots of the equation `3x^2-243=0`
APPEARS IN
संबंधित प्रश्न
In the following, determine whether the given quadratic equation have real roots and if so, find the roots:
3x2 - 2x + 2 = 0
Solve for x:
`1/x - 1/(x-2)=3`, x ≠ 0, 2
`11-x=2x^2`
`x^2-(sqrt3+1)x+sqrt3=0`
Find the nature of roots of the following quadratic equations:
`2x^2-8x+5=0`
If -4 is a root of the equation `x^2+2x+4p=0` find the value of k for the which the quadratic equation ` x^2+px(1+3k)+7(3+2k)=0` has equal roots.
If the roots of the quadratic equation `(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0` are real and equal, show that either a=0 or `(a^3+b^3+c^3=3abc)`
Find the values of k for which the given quadratic equation has real and distinct root:
`5x^2-kx+1=0`
If the roots of the equation `(a^2+b^2)x^2-2(ac+bd)x+(c^2+d^2)=0`are equal, prove that `a/b=c/d`
Solve for x: \[\frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6}, x \neq 3, - 5\]
