Question

2x^2 - 7x + 3. do by complete square

Answer 1

→  $2x^{2} -7x+3$

⇒ $2(x^{2} -\frac{7}{2} x+\frac{3}{2} )$

⇒ $2[(x-\frac{7}{4}) ^{2} -\frac{49}{16} +\frac{3}{2}]$

⇒ $2[(x-\frac{7}{4} )^{2} -\frac{25}{16}]$

⇒ $2[(x-\frac{7}{4} +\frac{5}{4} )(x-\frac{7}{4} -\frac{5}{4})]$

⇒ $2(x-\frac{1}{2} )(x-3)$

___________________________

❂ HOPE YOU GOT IT

Answer 2

Step-by-step explanation:

→ 2x^{2} -7x+32x

2

−7x+3

⇒ 2(x^{2} -\frac{7}{2} x+\frac{3}{2} )2(x

2

−

2

7

x+

2

3

)

⇒ 2[(x-\frac{7}{4}) ^{2} -\frac{49}{16} +\frac{3}{2}]2[(x−

4

7

)

2

−

16

49

+

2

3

]

⇒ 2[(x-\frac{7}{4} )^{2} -\frac{25}{16}]2[(x−

4

7

)

2

−

16

25

]

⇒ 2[(x-\frac{7}{4} +\frac{5}{4} )(x-\frac{7}{4} -\frac{5}{4})]2[(x−

4

7

+

4

5

)(x−

4

7

−

4

5

)]

⇒ 2(x-\frac{1}{2} )(x-3)2(x−

2

1

)(x−3)