Question

if one zero of the polynomial p(×) =×2-6×+k is twice the other then find the value of k

Answer 1

Solution

Given :-

• Polynomial , p(x) = x² - 6x + k
• one zeroes is twice the other

Find :-

• Find Value of k

Explanation

Let,

A/C to question,

• First zeroes be = p
• Second Zeroes be = 2p

Using Formula

★ Sum of zeroes = -b/a

★ product of zeroes = c/a

Where,

• a = coefficient of x²
• b = coefficient of x
• c = constant part

Then,

➡ Sum of zeroes = -(-6)/1

➡ p + 2p = 6

➡ 3p = 6

➡ p = 6/3

➡p = 2

Again,

➡ product of zeroes = k/1

➡ p * 2p = k

➡ 2p² = k

Keep value of p

➡ 2 * 2² = k

➡ k = 2 * 4

➡ k = 8

Hence

• Value of k will be = 8

________________

Answer 2

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• The polynomial p(x) = x² - 6x + k

• One zero of the polynomial is twice the other.

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The value of k.

$\bf{\underline{\underline\green{Solution:-}}}$

The polynomial = x² - 6x + k

Given polynomial is in the form ax² + bx + c

Here:-

• a = 1

• b = -6

• c = k

Let one zero of the polynomial be ' p ' and the other zero will be ' 2p'

As we know that:-

$\rm\longrightarrow Sum \: of \: zeroes \: = \dfrac{ - b}{ \: a}$

$\rm\longrightarrow Product\: of \: zeroes \: = \dfrac{ c}{ a}$

Now:-

$\rm\longrightarrow Product\: of \: zeroes \: = p \times 2p= \dfrac{c}{a}$

$\rm\longrightarrow p \times 2p= \dfrac{k}{1}$

$\rm\longrightarrow 2 {p}^{2} = {k} ......(i)$

$\rm\longrightarrow Sum\: of \: zeroes \: = p + 2p= \dfrac{-b}{a}$

$\rm\longrightarrow p + 2p= \dfrac{-(-6)}{1}$

$\rm\longrightarrow 3p= 6 ......(i)$

$\rm\longrightarrow {p}= \dfrac{ 6}{ \: \: 3}$

$\rm\longrightarrow {p}= 2 .......(ii)$

Now, Substitute p = 2 in equation (i)

$\rm\longrightarrow 2 {p}^{2} = {k}$

$\rm\longrightarrow 2 {(2)}^{2} = {k}$

$\rm\longrightarrow 2 \times 4 = {k}$

$\rm\longrightarrow k = 8$

Therefore:-

$\underline{\boxed{\bf \purple{\longrightarrow k = 8}} }$