Question

If p(x) = x3 + bx2 + cx + 5 has one zero
√5+2 and b and c are rational numbers
then find the sum of other two zeroes.
√5+7
√5-7
5+ √7
7-√5
​

Answer 1

Answer:

7 - √5

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Answer 2

The sum of other two zeroes = 7 - √5

Given :

p(x) = x³ + bx² + cx + 5 has one zero √5 + 2 and b and c are rational numbers

To find :

The sum of other two zeroes is

• √5 + 7

• √5 - 7

• 5 + √7

• 7 - √5

Solution :

Step 1 of 4 :

Write down first zero of the polynomial

Here it is given that p(x) = x³ + bx² + cx + 5 has one zero √5 + 2 and b and c are rational numbers

Step 2 of 4 :

Find second zero of the polynomial

Since irrational roots occurs in conjugate pair

So 2 - √5 is another zero of the polynomial

Step 3 of 4 :

Find third zero of the polynomial

The degree of the polynomial = 3

Third root must be real

Let the third zero is p

Then we have

$\displaystyle \sf{ Product \: of \: the \: zeroes = - \frac{5}{1} }$

$\displaystyle \sf{ \implies (2 + \sqrt{5}) \times (2 - \sqrt{5} ) \times p = - 5 }$

$\displaystyle \sf{ \implies \{ {2}^{2} - {( \sqrt{5} )}^{2} \} \times p = - 5 }$

$\displaystyle \sf{ \implies (4 - 5) \times p = - 5 }$

$\displaystyle \sf{ \implies ( - 1) \times p = - 5 }$

$\displaystyle \sf{ \implies p = 5 }$

Third zero of the polynomial = 5

Step 4 of 4 :

Find sum of other two zeroes

The sum of other two zeroes

= Second zero + Third zero

= 2 - √5 + 5

= 7 - √5

Hence the correct option is 7 - √5