Question

Find the quadratic polynomial whose zeroes are 3+ root5 and 3-root5 find all the zeroes of p(x)=x^4-5x^3-2x^3+10x-8 .If 2 of its zeroes are root 2-root2

Answer 1

Question:-

Find the quadratic polynomial whose zeroes are 3+√5 and 3-√5.

Solution:-

Let $\sf{\alpha}$ and $\sf{\beta}$ are the zeros of the polynomial.

$\sf{\alpha\:=\:3+\sqrt{5}}$ and $\sf{\beta\:=\:3-\sqrt{5}}$

Sum of zeros = $\sf{\alpha\:+\:\beta}$

⇒ 3 + √5 + 3 - √5

⇒ 6

Product of zeros = $\sf{\alpha \beta}$

⇒ (3 + √5)(3 - √5)

(a + b)(a - b) = a² - b²

⇒ (3)² -(√5)²

⇒ 9 - 5

⇒ 4

We know that, polynomial is given by

k [x² - (Sum of zeros)x + Product of zeros]

Where k = 1

⇒ x² - 6x + 4 is the required polynomial.

$\rule{100}2$

Question:-

Find all the zeroes of p(x) = x⁴ - 5x³ - 2x² +10x - 8. If two of its zeroes are √2, -√2.

Solution:-

(x + √2) and (x -√2) are the two zeros of the factor p(x).

(x + √2)(x - √2) = x² - 2 is the factor of p(x).

Now, divide the x² - 2 by x⁴ - 5x³ - 2x² + 10x - 8.

$\begin{minipage}{7 cm}\quad\begin{array}{m{3.5em}cccc} & & x^2 & -5x & +4 \\ \cline{1-6}\multicolumn{2}{l}{x^2-2\big)} & x^4 & -5x^3 & -2x^2 & +10x -8\\ & & -(x^4 & -0 & -2x^2) & & \\ \cline{3-4} & & & -5x^3 & +4x^2 & +10x & & & & -(5x^3 & +0 & +10x) & \\ \cline{4-5} & & & & 4x^2 & -8 \\ & & & & -(4x^2 & -8) \\ \cline{5-6} & & & & & 0 \\ \end{array}\end{minipage}$

So,

p(x) = (x² - 2) (x² - 5x + 4)

= (x² - 2) (x² - 4x - x + 4)

= (x² - 2) [x(x - 4) -1(x - 4)]

= (x + √2) (x - √2) (x - 4) (x - 1) are the zeros

Answer 2

Question (1) :---- Find the quadratic polynomial whose zeroes are (3+√5) and (3 - √5)...

Concept used :----

❁ if α and β are the roots of a quadratic eqn , the eqn can be written as. x²−(α+β)x+αβ = 0 ..

❁ or we can say that, the Equation will be x² -(sum of zeros ) + Product of zeros = 0 .

Solution :-----

➥ Given , α = (3+√5) and , β = (3-√5)

So,

➥ sum of zeros = (α+β) = (3+√5) + (3-√5) = 6

➥ Product of zeros = (α×β) = (3+√5)×(3-√5) = (3)² - (√5)² = 9 - 5 = 4 [ using (a-b)(a+b) = a² - b²

So, Our Quadratic Equation will be :-----

➥ x²−(α+β)x+αβ = 0 = x² - 6x + 4 (Ans) .

Hence, The Quadratic Polynomial wjose zeros are 3+√5 and 3-√5 is x² - 6x + 4 = 0 ..

______________________________

Question (2) :---- find all the zeroes of p(x)=x^4-5x^3+2x²+10x-8 .If 2 of its zeroes are √2 and (-√2) ...

Concept used :---- The zero of the polynomial is defined as any real value of x, for which the value of the polynomial becomes zero.

A real number k is a zero of a polynomial p(x), if p(k) = 0.

Solution :----

The Given Polynomial is === x^4-5x^3+2x²+10x-8

Now, √2 and (-√2) are two zeros of this polynomial ..

it Means that, (x-√2) and (x+√2) are the Factors of the given Polynomial ...

[ Now, we know that, if f(x) and g(x) are the factors of a polynomial , than f(x)*g(x) will also be the factor of same Polynomial .. ]

So,

→ (x-√2)(x+√2) = x² - 2 { (a-b)(a+b) = a² - b² }

So, we can say that, (x²-2) will be the Factor of given Polynomial ..

_______________________

Now, we will divide the given Polynomial to Find other Factors ,,, [ By long Division Method ]

x² - 2) x^4-5x^3+2x²+10x-8 ( x² - 5x + 4

(-) x^4. (+) -2x²

-5x³ + 4x² + 10x - 8

(+) -5x³. (-)+10x

4x² - 8

4x² - 8

0.

So, our Quotient will come = (x² - 5x + 4)

Equate this Equal to zero Now, to Find other To Factors .

_________________________

By, splitting the Middle term now, we get,

☛ (x² - 5x + 4) = 0

☛ x² - 4x - x + 4 = 0

☛ x(x-4) -1(x -4 ) = 0

Taking (x-4) common Now,

☛ (x-4)(x-1) = 0

Putting both Equal to zeros now, we get,

→ x - 4 = 0 and , x - 1 = 0

x = 4 x = 1 ...

Hence, we can say that, other two zeros of Given Polynomial are 4 and 1 ...