Question

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

Answer 1

Given :

• a and b are the zeroes of f(x) = x² - 5x + k .

• a - b = 1

To find :

• The value of k .

Solution :

• f (x) = x² -5x + k .

Here ,

• a = 1 , b = -5 and c = k .

• Sum of Zeroes ( a+b) = -b/a = 5 .

• product of zeroes (ab) = c/a = k .

$\implies \: a \: - b = 1 \\ \\ \implies \: \sqrt{ {(a + b)}^{2} - 4ab } = 1 \\ \\ \implies \: \sqrt{ {(5)}^{2} - 4(k) = 1} \\ \\ \implies \: 25 - 4k = 1 \\ \\ \implies \: 25 - 1 = 4k \\ \\ \implies \: 24 = 4k \\ \\ \implies \: k = \cancel{\frac{24}{4} } \\ \\ \implies \: k \: = 6$

Hence , the value of k is 6.

Answer 2

Answer:

$\boxed{Given} :$

• • a and b are the zeroes of f(x) = x² - 5x + k .

• • a - b = 1

To find :

• The value of k .

Solution :

• f (x) = x² -5x + k .

Here ,

• a = 1 , b = -5 and c = k .

• Sum of Zeroes ( a+b) = -b/a = 5 .

• product of zeroes (ab) = c/a = k .

x2-5x+k

Here, a=1, b=-5 and c=k

Now, α+ β = -b/a= -(-5)/1= 5

α*β = c/a= k/7= k

Now,α - β =1

Squaring both sides, we get,

(α - β)2=12

⇒ α2 + β2 - 2αβ = 1

⇒ (α2 + β2 + 2αβ) - 4αβ = 1

⇒ (α +β)2 -4αβ =1

⇒ (5)2-4k=1

⇒ -4k= 7-25

⇒ -4k= -24

⇒ k=6 So the value of k is 6.