Question

find the quadratic polynomial whose zeroes are 7 - 5√2and 7 + 5√2​

Answer 1

Given

Zeroes of Polynomial 7 - 5√2and 7 + 5√2​

To Find

quadratic polynomial

Now let

α = 7 - 5√2

β = 7 + 5√2​

General Equation of quadratic polynomial

x² - (α+β)x - αβ = 0

Sum of zeroes

( α + β) = (7 - 5√2+ 7 + 5√2​)

( α + β) = (7 + 7​)

( α + β) = (14​)

Product of zeroes

αβ = (7-5√2)(7+5√2)

αβ = 7² - (5√2)²

αβ = 49 - 25×2

αβ = 49 - 50

αβ = -1

Put the value on

x² - (α+β)x - αβ = 0

x² - 14x - (-1) = 0

x² - 14x + 1 = 0

Answer

x² - 14x + 1 = 0

Answer 2

Question:-

• Find the quadratic polynomial whose zeroes are 7 - 5√2and 7 + 5√2

To Find:-

• Find the quadratic equation.

Solution:-

Formula to be Used:-

• x² - ( à + ß )x + àß = 0

Substituting Values:-

$\sf\longrightarrow \: { x }^{ 2 } - ( \alpha + \beta )x + \alpha \beta = 0$

$\sf\longrightarrow \: { x }^{ 2 } - ( 7 - 5 \sqrt { 2 } + 7 + 5 \sqrt { 2 } )x + 7 - 5 \sqrt { 2 } \times 7 + 5 \sqrt { 2 } = 0$

$\sf\longrightarrow \: { x }^{ 2 } - ( 14 )x + { ( 7 ) }^{ 2 } - { ( 5 \sqrt { 2 } ) }^{ 2 }$

$\sf\longrightarrow \: { x }^{ 2 } - 14x + 49 - 50 = 0$

$\sf\longrightarrow \: { x }^{ 2 } - 14x - 1 = 0$

Hence ,

• Quadratic Equation is $\sf \: { x }^{ 2 } - 14x - 1 = 0$