Find the roots of the quadratic equation 3`x^2` − 7`x` − 6 = 0.
Given,
Hence, the roots of the equation (3`x^2` - 7`x` - 6 = 0) are 3 and `-2/3`.
3`x^2` - 7`x` - 6 = 0
Comparing with `ax^2` - `bx` + `c` = 0
So, a = 3, b = -7 and c = -6
We Know that,
D = `"b"^2` - 4ac
⇒ D = `(-7)^2` - 4`\times`3`\times` (-6)
⇒ D = 49 + 72
⇒ D = 121
So, the root of the equation is given by
⇒ x = `\frac{-b\pm\sqrt{D}}{2a}`
⇒ x = `\frac{-(-7)\pm\sqrt{121}}{2\times3}`
⇒ x = `\frac{7\pm11}{2\times6}`
Now,
| ⇒ x = `"7 + 11"/6″` ⇒ x = `18/6` ⇒ x = 3 | ⇒ x = `"7 – 11"/6` ⇒ x = `-4/6` ⇒ x = `-2/3` |