Question

2x2-5x+2=0 find the roots​

Answer 1

Answer:

The given quadratic equation is 2x

2

+5x+2=0

On comparing with ax

2

+bx+c=0

We get, a=2,b=5,c=2

∴ b

2

−4ac=(5)

2

−4×2×2

=25−16

=9

x=

2a

−b±

b

2

−4ac

=

2×2

−5±

9

=

4

−5±3

x=

4

−5+3

x=−

2

1

or x=−2

∴ −

2

1

and −2 are the roots of given quadratic equation.

Answer 2

Answer:

Given Quadratic polynomial : 2x² - 5x + 2

Finding Roots :-

${ \implies{ \sf{2 {x}^{2} - 5x + 2 = 0}}} \\ \\ { \implies{ \sf{2 {x}^{2} - 4x - 1x + 2 = 0}}} \\ \\ { \implies{ \sf{2x(x - 2) - 1(x - 2) = 0}}} \\ \\ { \implies{ \sf{(2x - 1)(x - 2) = 0}}} \\ \\ { \implies{ \sf{2x - 1 = 0 \: (or) \: x - 2 = 0}}} \\ \\ { \implies{ \sf{2x = 1 \: (or) \: x = 2}}} \\ \\ {\bf { \implies{ \bf{x = \frac{1}{2} \: or \: 2}}}}$

Therefore,

• Roots of polynomial = 1/2 or 2

Verification :

$: \: \: \to{ \sf{p(x) = 2 {x }^{2} - 5x + 2 }} \\ \\ : \: \: \to{ \sf{p(2) = 2 ({2)}^{2} - 5(2) + 2 }} \\ \\ : \: \: \to{ \sf{p(2) = 2(4) - 10 + 2}} \\ \\ : \: \: \to{ \sf{p(2) = 10 - 10}} \\ \\ : \: \: \to{ \sf{p(2) = 0}}$

$\: : \: \: \to{ \sf{p \bigg( \frac{1}{2} \bigg) = 2 { \bigg(\frac{1}{2} \bigg) }^{2} - 5 \bigg( \frac{1}{2} \bigg) + 2 }} \\ \\ : \: \: \to \sf{p \bigg( \frac{1}{2} \bigg) = \frac{2}{4} - \frac{5}{2} + 2 } \\ \\ : \: \: { \to{ \sf{p \bigg( \frac{1}{2} \bigg) = \frac{2 - 10}{4} + 2}}} \\ \\ : \: \: { \to{ \sf{p \bigg( \frac{1}{2} \bigg) = \frac{ - 8}{4} + 2 }}} \\ \\ : \: \: { \to{ \sf{p \bigg( \frac{1}{2} \bigg) = - 2 + 2}}} \\ \\ : \: \: \to \sf{p \bigg( \frac{1}{2} \bigg) = 0}$

Hence Proved

${ \therefore \: { \sf{ \red{Roots \: of \: the \: polynomial =2 , \frac{1}{2} }}}}$