Question

lf alpha and beta are the zero of the quadratic polynomial Rx) =x^2+x-2, find a polynomial whose zeros are 1upon alpha minus 1upon beta

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Answer 1

AnsWer :

± 3 / 2.

SolutioN :

We have, Polynomial.

$\tt \dagger \: \: \: \: \: {x}^{2} + x - 2.$

Compare With General Expression.

ax² + bx + c = 0.

Where as,

• a = 1.
• b = 1.
• c = - 2.

$\rule{90}2$

→ x² + x - 2.

→ x² + 2x - x - 2.

→ x( x + 2 ) - 1( x + 2 )

→ ( x - 1 )( x + 2 )

Either,

→ x - 1 = 0.

→ x = 1.

Or,

→ x + 2 = 0.

→ x = - 2.

★ Sum of Zeros.

$\tt \dagger \: \: \: \: \: \alpha + \beta = \dfrac{ - b}{a}$

→ - b / a.

→ - 1.

★ Product Of Zeros.

$\tt \dagger \: \: \: \: \: \alpha \beta = \dfrac{ c}{a}$

→ c / a.

→ - 2.

★ Now, Let's Find.

$\tt \dashrightarrow \dfrac{1}{ \alpha } - \dfrac{1}{ \beta }$

$\tt \dashrightarrow \dfrac{ \alpha - \beta }{ \alpha \beta }$

#We know,

• ( a - b )² = ( a + b )² - 4ab

$\tt \dashrightarrow \dfrac{ {\alpha - \beta }^2}{ \alpha \beta }$

$\tt \dashrightarrow \dfrac{{ \alpha + \beta }^2- 4\alpha \beta }{ \alpha \beta }$

$\tt \dashrightarrow \dfrac{ 1 - 4( -2 )}{ - 2 }$

$\tt \dashrightarrow \dfrac{ 1 + 8}{ - 2 }$

$\tt \dashrightarrow \dfrac{ 9}{ - 2 }$

$\tt \dashrightarrow \dfrac{ \alpha -\beta = \pm3}{ - 2 }$

$\tt \dashrightarrow \dfrac{ \pm 3}{ - 2 }$

$\tt \dashrightarrow \pm \dfrac{ 3}{2 }$

Therefore, the required answer is ± 3 / 2.

Answer 2

Answer:

x²-3x+2=0

=x²-x-2x+2=0

=x(x-1)-2(x-1)=0

(x-1)(x-2)=0x=1

or ,

x=2 given zeroes of quadratic polynomial 1/2alpha+beta

1/2beta+alpha substitute=1/2(1)+2

1/2(2)+1=1/4

1/5formula is x²-(alpha+beta)x+alpha*beta =0therefore x²-

(1/4+1/5)x+1/4*1/5=0=ans is 20x^2 - 9x +1 =0