Question

20. If a and B are the zeroes of the quadratic polynomial f (x) = x2 – 2x + 3, find a polynomial whose roots are a. +2,
B +2.​

Answer 1

Step-by-step explanation:

The given polynomial

p(x) = x² - 2x + 3

Sum of the roots, A + B = 2

Product of the roots, AB = 3

Now, the polynomial with A + 2 and B + 2 as zeros is,

Sum of the roots, A + 2 + B + 2

= A + B + 4

= 2 + 4

= 6

Product of the roots, (A+2)(B+2)

= AB + 2A + 2B + 4

= AB + 2(A+B) + 4

= 3 + 2(2) + 4

= 11

The polynomial is in the structure,

p(x) = x² - (Sum of the roots)x + (Product of the roots)

p(x) = x² - 6x + 11

Answer 2

Question : -

If α & β are the zeroes of the quadratic polynomial f ( x ) = x² - 2x + 3 , find a polynomial whose roots are α + 2 , β + 2 .

Answer : -

Given : -

If α & β are the zeroes of the quadratic polynomial f ( x ) = x² - 2x + 3

Required to find : -

• find a polynomial whose roots are α + 2 , β + 2 .

Formula used : -

Quadratic formula ;

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

Solution : -

If α & β are the zeroes of the quadratic polynomial f ( x ) = x² - 2x + 3

we need to find a polynomial whose roots are α + 2 , β + 2 .

Here,

First let's find the roots of the given quadratic equation i.e. f ( x ) = x² - 2x + 3

For this let's use the quadratic formula ;

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

However,

The standard form of the quadratic equation is ax² + bx + c

Compare the given polynomial with the standard form of the polynomial .

Hence

• a = 1

• b = - 2

• c = 3

Substituting these values in the quadratic formula ;

$\tt x = \dfrac{ - ( - 2 ) \pm \sqrt{ {(-2)}^{2} - 4 ( 1 ) ( 3 )} }{ 2( 1 ) } \\ \\ \tt x = \dfrac{ + 2 \pm \sqrt{ ({ - 2)}^{2} - 4 (3)} }{2} \\ \\ \tt x = \dfrac{ + 2 \pm \sqrt{4 - 12} }{2} \\ \\ \tt x = \dfrac{ + 2 \pm \sqrt{ - 8} }{2} \\ \\ \tt x = \dfrac{2 \pm \sqrt{ - 2 \times 4} }{2} \\ \\ \tt x = \dfrac{2 \pm 2 \sqrt{ - 2} }{2} \\ \\ \sf{ This \: \: implies \: ;} \\ \\ \rm x = \frac{2 + 2 \sqrt{ - 2} }{2} \quad (and) \quad x = \frac{2 - 2 \sqrt{ - 2} }{2} \\ \\ \rm x = \dfrac{ 2(1 + \sqrt{ - 2} )}{2} \quad (and) \quad x = \frac{2(1 - \sqrt{ - 2} )}{2} \\ \\ \tt x = 1 + \sqrt{ - 2} \quad ( and) \quad x = 1 - \sqrt{ - 2} \\ \\ \sf hence \\ \\ \sf \alpha = 1 + \sqrt{ - 2} \quad ( and) \quad \beta = 1 - \sqrt{ - 2}$

Now,

Let's find the value of α + 2 , β + 2 .

Since,

α = 1 + √-2 , β = 1 - √-2

α + 2 =

=> 1 + √-2 + 2

=> 3 + √-2

β + 2 =

=> 1 - √-2 + 2

=> 3 - √-2

Hence,

α + 2 = 3 + √-2 ( and ) β + 2 = 3 - √-2

Now,

Let's frame the polynomial whose roots are α + 2 , β + 2

we know that ;

The formula to form a quadratic equation is ;

g ( x ) = x² - ( sum of the roots )x + product of the roots

g ( x ) = x² ( α + 2 + β + 2 ) x + α + 2 . β + 2

This implies ;

we need to find the values of α + 2 + β + 2 , α + 2 . β + 2

So,

α + 2 + β + 2 =

=> 3 + √-2 + 3 - √-2

=> + √-2 , - √-2 get's cancelled

=> 3 + 3

=> 6

α + 2 . β + 2 =

=> ( 3 + √-2 ) ( 3 - √-2 )

=> 3 ( 3 - √-2 ) + √-2 ( 3 - √- 2 )

=> 9 - 3 √-2 + 3 √- 2 + 2

=> 9 + 2

=> 11

Hence,

α + 2 + β + 2 = 6 , α + 2 . β + 2 = 11

Substitute this Values in the formula ;

g ( x ) = x² - ( α + 2 + β + 2 ) x + α + 2 . β + 2

g ( x ) = x² - ( 6 ) x + ( 11 )

g ( x ) = x² - 6x + 11

Therefore,

Required polynomial = x² - 6x + 11