Question

if one zero of the polynomial
p(x)= x²-6x +k is twice the other then find the value of k
​

Answer 1

Answer:

k=8

Step-by-step explanation:

The polynomial is p(x)= x²-6x +k

Let the zeros be 'x' and '2x' respectively.

• Sum of zeros = $\frac{-b}{a}$

put the values in above formula.

$x+2x=\frac{-(-6)}{1}$

$3x=6\\\therefore x=2$

• Product of Zeros = $\frac{c}{a}$

$2x \times x= \frac{k}{1}$

put the value of x in above formula.

$2\times2\times 2=k\\\therefore k=8$

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Answer 2

k = 8.

• The points at which a polynomial equals zero are known as its zeros. In layman's terms, we can state that a polynomial's zeros are variable values at which the polynomial equals 0. The zeros of a polynomial are often referred to as the equation's roots and are frequently.

here, according to the given information, we are given that,

The polynomial is,

p(x) = $x^{2} -6x +k$.

Now, we are given that, one zero of the polynomial is twice of the other zero of the given polynomial.

Then, let these zeros be y and 2y respectively.

Now, we know that the sum of the zeros is $\frac{-b}{a}$.

Then, we get, $y +2y= -\frac{-6}{1}$

Or, 3y = 6

Or, y = 2.

Now, we know that, the product of zeros is given as, $\frac{c}{a}$.

Then, we get,

$2y^{2} = \frac{k}{1}$

Now, since y = 2, putting this value, we get,

k = 8.

Hence, k = 8.

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