Question

if the product of zeroes of the polynomialx^2-9x+a is 8 then its zero are​

Answer 1

Given :-

• Product of Zeros of the polynomialx^2-9x+a = 8 .

To Find :-

• Zeros of the given Polynomial ?

Concept used :-

→ The sum of the roots of the Equation ax² + bx + c = 0 , is given by = (-b/a)

and ,

→ Product of roots of the Equation is given by = c/a.

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Solution :-

comparing x² - 9x + a = 0 with ax² + bx + c = 0, we get,

→ a = 1

→ b = (-9)

→ c = a

So,

→ Product of zeros = (c/a)

Putting values ,

→ (a/1) = 8

→ a = 8 .

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So, Given Quadratic Equation is x² - 9x + 8 = 0

→ x² - 9x + 8 = 0

Splitting The middle Term now,

→ x² - 8x - x + 8 = 0

→ x(x - 8) - 1(x - 8) = 0

→ (x - 8)(x - 1) = 0

→ x = 8 & 1 .

Hence, Zeros of the given Polynomial are 1 & 8.

Answer 2

Given :-

• Polynomial → x² - 9x + a

• Product of zeros → 8

To find :-

• The zeros of polynomial.

Solution :-

If the equation is ax² + bx + c = 0

Then the product of roots will be equal to $\frac{c}{a}$

So in the given polynomial a = 1 , b = - 9 , c = a

→ Product of zeros = 8

→ $\frac{c}{1}$ = 8

→ $\boxed{c \: = \: 8}$

Now the polynomial is x² - 9x + 8 .

For finding its zeros we will split its middle term .

→ x² - 9x + 8

→ x² - 8x - x + 8

→ x ( x - 8 ) -1 ( x -8 )

→ (x - 8) . ( x-1 )

→ x = 8 or x = 1

$\boxed{ Zeros\: = 1 \: , \: 8 }$

Verification

Putting x = 8

x² - 9x + 8 → 8² - 9(8) +8

→ 64 - 72 + 8

→ 72 - 72 = 0

Putting x = 1

x² - 9x +8 → 1² - 9(1) + 8

→ 1 - 9 + 8

→ 9-9 = 0

Hence verified .