Question

Find the quadratic polynomial whose zeroes are 2 root 7 and -5 root 7 .

Answer 1

Here , we have to find the required quadratic polynomial -

The given Zeroes are 2√7 and -5 √7

Sum of the zeroes -

=> ( 2√7 - 5√7)

=> ( -3 √ 7 )

Product Of Zeroes -

=> ( 2 √ 7 ) × ( -5 √ 7 )

=> -70

Now , Any Quadratic Polynomial can be written as -

=> x² - ( Sum of Zeroes ) x + ( Product Of Zeroes )

=> x² - ( -3 √ 7 )x + ( -70 )

=> x² + 3√7 x - 70

Thus , the required Quadratic Polynomial becomes x² + 3√7x - 70.

This is the answer .

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AddiTiOnaL InFoRmAtIon -

For a quadratic polynomial , ax² + bx + c

• Sum of roots = ( -b / a )

• Product of roots = ( c / a )

$\sf{ ( a + b )^n = ^n C_0 a ^ n b ^ 0 + ^n C_1 a ^ {n - 1 } b ^ 1 + ... + ^n a^ 0 b ^ n }$

$\sf{ ( a - b )^n = ^n C_0 a ^ n b ^ 0 - ^n C_1 a ^ {n - 1 } b ^ 1 + ... - ^n a^ 0 b ^ n }$

(x - y )( x + y ) = x² - y²

x⁴ + x² y² + y⁴ { Sophie German identity }

=> x⁴ + 2x²y² + y⁴ - x²y²

=> ( x² + y² )² - (xy)²

=> (x² + xy + y²)( x² -xy + y² )

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Answer 2

★ Given :

roots are -

$\alpha = 2\sqrt{7} \\ \\ \\ \beta = -5\sqrt{7}$

★ To find :

The quadratic polynomial whose roots are 2√7 and -5√7.

★ We know that,

form of quadratic equation -

${\pink{\boxed{x^2 - (\alpha + \beta)x + (\alpha \times \beta)}}}\\ \\ \\ where, \alpha\: and \: \beta \:are \:roots$

★ Solution :

by putting the values of roots ,

$x^2 - [2\sqrt{7} + (-5\sqrt{7})]x + (2\sqrt{7} \times -5\sqrt{7}) \\ \\ \\ x^2 - (-3\sqrt{7})x + (-70) \\ \\ \\ \bold{x^2 + 3\sqrt{7} - 70}$

★ Verification:

f(x) = x² + 3√7x -70

f(2√7) = (2√7}² + 3√7(2√7) - 70

= 28 + 42 - 70

= 70 - 70

= 0

therefore it is a root.

f(-5√7) = (-5√7)² + 3√7(-5√7) - 70

= 175 - 105 - 70

= 175 - 175

= 0

therefore it is also a root.

hence , x²+3√7 x - 70 has two roots 2√7 and -5√7 .