Question

8u^2-16u. Find zeroes of this polynomial

Answer 1

AnsWer :

u = 0 and u = 2.

Solution :

We have, Polynomial.

$\tt \dagger \: \: \: \: \: 8 {u}^{2} - 16u$

Compare With General Formula.

$\tt \dagger \: \: \: \: \: a {x}^{2} + bx + c.$

Taking, Expression.

$\tt : \implies 8 {u}^{2} - 16u = 0.$

$\tt : \implies 8 (u- 2) = 0.$

$\tt : \implies 8 = 0.$

Or,

$\tt : \implies u - 2 = 0.$

$\tt : \implies u = 2.$

★ Let try With Different method ( Quadratic Formula )

$\tt \dagger \: \: \: \: \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

Where as,

• a = 8.
• b = 16.
• c =0.

$\tt : \implies x = \dfrac{ 16 \pm \sqrt{ {16 }^{2} - 4 \times 8 \times 0 } }{16}$

$\tt : \implies x = \dfrac{ 16 \pm\sqrt{ 256 } }{16}$

$\tt : \implies x = \dfrac{ 16 \pm 16 }{16}$

Either,

$\tt : \implies x = \dfrac{ 16 + 16 }{16}$

$\tt : \implies x = \dfrac{ 32 }{16}$

$\tt : \implies x = 2.$

Or,

$\tt : \implies x = \dfrac{ 16 - 16 }{16}$

$\tt : \implies x = \dfrac{ 0 }{16}$

$\tt : \implies x = 0.$