Question

If α and β are the zeros of the polynomial f (x) = x² - 5x + k such that α - β = 1, find the value of k

Answer 1

Answer:

k = 6

Step-by-step explanation:

Given Polynomial = x² – 5x + k

And, α – β = 1

Value of k = ??

$\rm{ \alpha - \beta = 1 --- (Equation\:I)}$

According to the relationship between zeros and polynomial's zeros,

$\rm{ \alpha + \beta = \dfrac{ - b}{a}}$

Where, a is 1 and b is –5.

$\rm{\rightarrow{\alpha + \beta = \dfrac{ - ( - 5)}{1}}}$

$\rm{\rightarrow{\alpha + \beta = 5}}$

$\rightarrow{\rm{\alpha =5 - \beta \: --- (Equation \: II)}}$

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Substitute equation II in equation I,

$\rm{\rightarrow{\alpha - \beta = 1}}$

$\rm{\rightarrow{(5 - \beta ) - \beta = 1}}$

$\rm{\rightarrow{5 - 2\beta = 1}}$

$\rm{\rightarrow{ - 2\beta = 1 - 5}}$

$\rm{\rightarrow{ - 2\beta = - 4}}$

$\rm{\rightarrow{ \beta = \dfrac{ - 4}{ - 2}}}$

$\rm{\rightarrow{\beta =2}}$

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Substitute the value of $\beta$ in equation I,

$\rm{\rightarrow{\alpha - \beta = 1}}$

$\rm{\rightarrow{\alpha - 2 = 1}}$

$\rm{\rightarrow{\alpha = 1 + 2}}$

$\rm{\rightarrow{ \alpha = 3}}$

The zeros of polynomial are 3 and 2.

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A polynomial is formed as :

$\rightarrow$ x² – (Sum of zeros)x + Product of zeros

In the given polynomial, x² - 5x + k the product of zeros is the value of k.

$\rightarrow$ x² – (3 + 2)x + (3 × 2)

$\rightarrow$ x² – 5x + 6

$\rightarrow$ k = 6

Answer 2

Answer:

Given :-

• If α and β are the zeros of the polynomial f(x) = x² - 5x + k such that α - β = 1.

To Find :-

• What is the value of k.

Formula Used :-

$\clubsuit$ Sum of roots :

$\longmapsto \sf\boxed{\bold{\pink{\alpha + \beta =\: \dfrac{- b}{a}}}}$

$\clubsuit$ General Formula For Quadratic Equation :

$\longmapsto \sf\boxed{\bold{\pink{x^2 - (\alpha + \beta)x + \alpha\beta =\: 0}}}$

where,

• α + β = Sum of roots
• αβ = Product of roots

Solution :-

Given equation :

$\bigstar \: \: \sf\bold{x^2 - 5x + k}$

where,

• a = 1
• b = - 5
• c = k

Then,

$\implies \sf \alpha - \beta =\: 1$

$\implies \sf\bold{\purple{\alpha - \beta =\: 1\: ------\: (Equation\: No\: 1)}}$

Again, by using the formula we get,

$\implies \sf \alpha + \beta =\: \dfrac{- (- 5)}{1}$

$\implies \sf \alpha + \beta =\: \dfrac{5}{1}$

$\implies \sf \alpha + \beta =\: 5$

$\implies \sf \alpha =\: 5 - \beta$

$\implies \sf \bold{\purple{\alpha =\: 5 - \beta\: ------\: (Equation\: No\: 2)}}$

Now, by putting the value of α in the equation no 1 we get,

$\implies \sf (5 - \beta) - \beta =\: 1$

$\implies \sf 5 - 2\beta =\: 1$

$\implies \sf 5 - 1 =\: 2\beta$

$\implies \sf 4 =\: 2\beta$

$\implies \sf \dfrac{\cancel{4}}{\cancel{2}} =\: \beta$

$\implies \sf \dfrac{2}{1} =\: \beta$

$\implies \sf 2 =\: \beta$

$\implies \sf\bold{\green{\beta =\: 2}}$

Again, by putting β = 2 in the equation no 2 we get,

$\implies \sf \alpha =\: 5 - \beta$

$\implies \sf \alpha =\: 5 - 2$

$\implies \sf \bold{\green{\alpha =\: 3}}$

Hence, we get :

• α = 3
• β = 2

Now, we have to find the value of k ,

Given :

$\bigstar \: \bold{x^2 - 5x + k}$

According to the question by using the formula we get,

$\implies \sf x^2 - (\alpha + \beta)x + \alpha\beta =\: 0$

By putting we get,

• α = 3
• β = 2

$\implies \sf x^2 - (3 + 2)x + 3 \times 2 =\: 0$

$\implies \sf x^2 - 5x + 6 =\: 0$

$\implies \sf k =\: 6$

$\implies \sf\bold{\red{k =\: 6}}$

$\therefore$ The value of k is 6 .