Question

If α,β are the roots of the quadratic equation ax^2+bx+c=0, find quadratic equation whose roots are α^2+β^2 and α^-2-β^-2

Answer 1

Answer:

α² + ß² = b²/a² - 2c/a

α^(-2) + ß^(-2) = b²/c² - 2a/c

Note:

★ The possible values of the variable for which the polynomial becomes zero are called its zeros.

★ To find the zeros of the polynomial p(x) , operate on p(x) = 0 .

★ A quadratic polynomial can have atmost two zeros .

★ If α and ß are the zeros of the quadratic polynomial ax² + bx + c , then ;

• Sum of zeros , (α + ß) = -b/a

• Product of zeros , (αß) = c/a

★ If α and ß are the zeros of any quadratic polynomial , then it is given by ;

x² - (α + ß)x + αß

★ If α and ß are the zeros of the quadratic polynomial ax² + bx + c , then they (α and ß) are also the zeros of the quadratic polynomial k(ax² + bx + c) , k≠0.

Solution:

Here,

The given quadratic polynomial is ;

ax² + bx + c .

Also,

It is given that , α and ß are the zeros of the given quadratic polynomial .

Thus,

Sum of zeros will be ;

α + ß = -b/a

Also,

Product of zeros will be ;

αß = c/a

1) α² + ß² = (α + ß)² - 2αß

= (-b/a)² - 2(c/a)

= b²/a² - 2c/a

2) α^(-2) + ß^(-2) = 1/α² + 1/ß²

= (ß² + α²) / α²ß²

= (α² + ß²) / (αß)²

= [ (α + ß)² - 2αß ] / (αß)²

= (α + ß)²/(αß)² - 2αß/(αß)²

= (-b/a)²/(c/a)² - 2/αß

= (b²/a²)/(c²/a²) - 2/(c/a)

= (b²/a²)×(a²/c²) - 2×(a/c)

= b²/c² - 2a/c

Hence,

α² + ß² = b²/a² - 2c/a

α^(-2) + ß^(-2) = b²/c² - 2a/c