- Home
- Standard 11
- Mathematics
4-2.Quadratic Equations and Inequations
easy
If the sum of the two roots of the equation $4{x^3} + 16{x^2} - 9x - 36 = 0$ is zero, then the roots are
A
$1, 2 -2$
B
$ - 2,\frac{2}{3}, - \frac{2}{3}$
C
$ - 3,\frac{3}{2}, - \frac{3}{2}$
D
$ - 4,\frac{3}{2}, - \frac{3}{2}$
Solution
(d) Given equation $4{x^3} + 16{x^2} – 9x – 36 = 0$,
Putting $x = – 4$ ==> $ – 4 \times 64 + 256 + 36 – 36 = 0$
Hence $x = – 4$ is a root of the equation
Now reduced equation is $4{x^2}(x + 4) – 9(x + 4) = 0$
==> $(x + 4)(4{x^2} – 9) = 0$ ==> $x = – 4,x = \pm \frac{3}{2}$
Thus roots are $ – 4\,, – \frac{3}{2},\frac{3}{2}$
Standard 11
Mathematics
