Question

if two zeros of a polynomial 2x^4-2x^3-7x^2+3x+6 are -√5/2 and +√3/2, find other two zeros

Answer 1

Correct QuestioN :

If two zeros of a polynomial 2x⁴ -2x³ - 7x² + 3x + 6 are √3/2 and - √3/2, Find other two zeros ?

SolutioN :

Let,

• The Zero be α and β.

$\tt \dagger \: \: \: \: \: \alpha = \sqrt{ \dfrac{3}{2} } \: \: \: and \: \: \: \beta = - \sqrt{ \dfrac{3}{2} }$

We know,

$\tt \dagger \: \: \: \: \: Sum \: Of \: Zeros. \\ \tt \alpha + \beta = \frac{-b}{a} = \frac{Coefficient \: of \: x}{ Coefficient \: of \: x^2 }$

$\tt \dagger \: \: \: \: \: Product \: Of \: Zeros. \\ \tt \alpha \times \beta = \frac{c}{a} = \dfrac{Constant \: terms}{ Coefficient \: of \: x^2}$

$\rule{100}2$

So,

☛ Sum of Zero.

$\tt : \implies \alpha + \beta = - \sqrt{\dfrac{3}{2} } + \sqrt{ \dfrac{3}{2} }$

$\tt : \implies \alpha + \beta = \sqrt{\dfrac{3}{2} } - \sqrt{ \dfrac{3}{2} }$

$\tt : \implies \alpha + \beta = \sqrt{\dfrac{3 - 3}{2} }$

$\tt : \implies \alpha + \beta = 0.$

$\rule{170}2$

☛ Product of Zeros.

$\tt : \implies \alpha \times \beta = \sqrt{\dfrac{3 }{2} } \times - \sqrt{\dfrac{3}{2}}$

$\tt : \implies \alpha \times \beta = - \dfrac{3 }{2}$

$\rule{170}2$

✎ Now, Putting the value,

$\tt \dagger \: \: \: \: \: k\Big[{x}^{2} - Sx + P\Big]$

Where as,

• K Constant term.
• S Sum of Zero.
• P Product of Zero.

$\tt : \implies k\Bigg[{x}^{2} - \big(0 \big)x + \bigg( - \dfrac{3}{2} \bigg)\Bigg]$

$\tt : \implies k\Bigg[{x}^{2} - \dfrac{3}{2} \Bigg]$

$\tt : \implies k\Bigg[ \dfrac{2 {x}^{2} - 3}{2} \Bigg]$

$\tt : \implies k\Bigg[ {2 {x}^{2} - 3} \Bigg]$

$\rule{170}2$

So, When you divided by 2x² - 5. We get.

2x² - 3 ) 2x⁴ - 2x³ - 7x² + 3x + 6 ( x² - x - 2

................2x⁴............-3x²

________________________________

.,....,....,............ -2x³ - 4x² + 3x + 6

..,.............,........- 2x³ ........ + 3x

___________________________________

.....................................- 4x .......... + 6

..................................... -4x .......... + 6

___________________________________

............................................................0.

Taking Q → x² - x - 2.

$\tt : \implies {x}^{2} - x - 2 = 0.$

$\tt : \implies {x}^{2} - 2x + x- 2 = 0.$

$\tt : \implies x(x - 2) + 1(x - 2) = 0.$

$\tt : \implies (x - 2)( x+ 1) = 0.$

Either,

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

$\tt : \implies x = 2.$

Or,

$\tt : \implies x+ 1 = 0.$

$\tt : \implies x = - 1$

✡ All Zeros is √3 / 2 , - √3/ 2 , 2 , - 1.

Answer 2

YouR AnsweR RefeR To ThE AttachmenT