Question

If the roop of a quadratic equation are (3 +root 5) and (3-root5). Then form the quadratic equation
​

Answer 1

Answer:

x² - 6x + 4

Step-by-step explanation:

Given : Two roots of quadratic equation : (3 + √5) and (3 - √5)

Let the roots of required quadratic equation be α and β.

α = 3 + √5 and β = 3 - √5

Now,

Sum of zeroes = α + β

= (3 + √5) + (3 - √5)

= 6

Product of zeroes = αβ

= (3 + √5)(3 - √5)

Identity : (a + b)(a - b) = a² - b²

Here, a = 3, b = √5

= (3)² - (√5)²

= 9 - 5

= 4

The required quadratic equation is :

→ p(x) = k [x² - (α + β)x + αβ]

Putting known values, we get

→ p(x) = k [x² - (6)x + 4]

→ p(x) = k [x² - 6x + 4]

Put k = 1, we get

→ p(x) = x² - 6x + 4

Answer 2

$\bf{\Huge{\boxed{\rm{\blue{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

If the roots of a quadratic equation are (3+√5) & (3-√5).

$\bf{\Large{\underline{\bf{To\:find\::}}}}$

The form of quadratic equation.

$\bf{\Large{\underline{\tt{\red{Explanation\::}}}}}$

$\bf{We\:have}\begin{cases}\rm{\alpha \:=\:(3+\sqrt{5} )}\\\\ \rm{\beta \:=\:(3-\sqrt{5} )}\end{cases}}$

Now,

$\bf{\Large{\boxed{\sf{Sum\:of\:the\:root\::}}}}}$

$\longmapsto\sf{\alpha \:+\:\beta }$

$\longmapsto\sf{(3+\sqrt{5} )\:+\:(3-\sqrt{5} )}$

$\longmapsto\sf{3\cancel{+\sqrt{5} }+3\cancel{-\sqrt{5} }}$

$\longmapsto\sf{\red{6}}$

$\bf{\Large{\boxed{\sf{Product\:of\:root\::}}}}}$

$\longmapsto\sf{\alpha .\beta }$

$\longmapsto\sf{(3+\sqrt{5} )(3-\sqrt{5} )}$

$\longmapsto\sf{(3)^{2} -(\sqrt{5} )^{2} }$

$\longmapsto\sf{9\:-\:5}$

$\longmapsto\sf{\red{4}}$

Therefore,

$\implies\rm{Required\:quadratic\:polynomial\:is\:\Large{\boxed{x^{2} -(\alpha +\beta )x+\alpha \beta }}}}}}$

$\implies\rm{x^{2} -(6)x\:+\:(4)}$

$\implies{\rm{\red{x^{2}\: -\:6x\:+\:4}}}$

_____________________________________________

$\mathbb{\underline{VERIFICATION\::}}$

$\mapsto\sf{Sum\:of\:roots\:=\:3+\cancel{\sqrt{5}} \:+\:3\cancel{-\sqrt{5} }}$

$\mapsto\sf{Sum\:of\:roots\:=\:6\:=\frac{6}{1} =\frac{-Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }$

&

$\mapsto\sf{Product\:of\:roots\:=\:(3+\sqrt{5} )(3-\sqrt{5} )}$

$\mapsto\sf{Product\:of\:roots\:=\:9\:-\:5}$

$\mapsto\sf{Product\:of\:roots\:=\:4\:=\:\frac{4}{1} \:=\;\frac{Constant\:term}{Coefficient\:of\:x^{2} } }$