Question

please solve the question please ​

Answer 1

Answer:

k = 3

Explanation:

Given polynomial,

$\rm \implies \: f(x) = {x}^{2} - 4x + k$

$\rm \: Whose \: zeros \: \alpha \: and \: \beta$

We need to find the value of ‘k’

On comapring f(x) with the general form of a quadratic equation ax² + bx + c

We get,

• a = 1
• b = -4
• c = k

Then,

$\rm \bullet \: sum \: of \: zeros = \alpha + \beta = \dfrac{ - b}{a} = \dfrac{ - ( - 4)}{1} = 4$

$\rm \bullet \: product \: of \: zeros = \alpha \beta = \dfrac{c}{a} = \dfrac{k}{1} = k$

Given,

$\implies \: \alpha - \beta = 2$

On squaring both sides,

$\rm \implies \: {( \alpha - \beta )}^{2} = {(2)}^{2}$

$\rm \implies \: {( \alpha + \beta) }^{2} - 4 \alpha \beta = 4$

Substituting the values,

$\rm \implies \: {( 4) }^{2} - 4 k = 4$

$\rm \implies \: 16- 4 k = 4$

$\rm \implies \: 16- 4 = 4 k$

$\rm \implies \: 12 = 4 k$

$\rm \implies \: \dfrac{12}{4} = k$

$\rm \implies \: 3 = k$

∴ The value of ‘k’ is 3

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Note:

$\bullet \: {( \alpha - \beta )}^{2} = {( \alpha + \beta) }^{2} - 4 \alpha \beta$