Question

hello mates.. ❤

find the zeroes of the following quadratic polynomial and verify the relationship between zeroes and its coefficents.

${x}^{2} - 49$

step by step explanation.

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Answer 1

X^2 - 49 = 0

x^2 - 7^2 = 0

As we know a^2 - b^2 = ( a+b)( a-b)

So

x^2 -7^2 = 0

( x+7)( x-7) = 0

x+7 = 0. or x-7 = 0

x= -7 , x= 7

Now let's verify

x^2 -49 = 0

here x= 7, -7

we know x^2 gives only positive quantity so it doesn't matter whether x is 7 or -7 ,it will give 49 only

So ( 7)^2 -49 = 49-49 = 0

(-7)^2 -49 = 49-49 = 0

Now another verification

Sum of zeroes = -( coefficient of x)/coefficient of x^2

So 7-7 =0

and coefficient of x= 0

Now product of zeroes = constant term/coefficient of x^2

Here constant term=-49
coefficient of x^2 = 1

So 7 × -7 = -49

and RHS = -49

so verified

✌✌✌❤❤Dhruv❤❤✌✌✌✌

Answer 2

▶ Question :-

→ Find the zeroes of the following quadratic polynomial and verify the relationship between zeroes and its coefficents.

${x}^{2} - 49$ .



▶ Answer :-

→ The zeros of f(x) are -7 and 7


▶ Step-by-step explanation :-


We have,

→ f(x) = ( x² - 49 ) .

= ( x² - 7² ) .

= ( x + 7 ) ( x - 7 ) .

[ °•° a² - b² = ( a + b )( a - b ) . ]


•°• f(x) = 0 .

==> ( x + 7 )( x - 7 ) = 0 .

==> x + 7 = 0 or x - 7 = 0 .

==> x = -7 or x = 7 .



So, the zeros of f(x) are -7 and 7 .


▶Now, VERIFICATION :-


→ Sum of zeros = - 7 + 7 = 0 = 0/1 = -( coefficient of x)/( coefficient of x² ) .



→ Product of zeros = -7 × 7 = -49 = -49/1 = ( constant term )/( coefficient of x² ) .



✔✔ Hence, it is solved and verified ✅✅ .




THANKS