Question

if the zeroes a and b of a polynomial x2-7x+k are such that a-b=1 then find the value of 'k '.​

Answer 1

Answer:

Polynomial p(x) = x2 -5x + k.

a – b = 1 ………..(i) (Given)

Sum of the zeroes,

a + b = 5 ……………..(ii)

Solving (i) and (ii), we get

a= 3 and b = 2.

Now product of the zeroes is given by,

ab= (constant term)/ (coefficient of x2)

ab= k

k = 3 × 2

k = 6.

Answer 2

Given:

• Polynomial => x²-7x + k
• a - b = 1
• a and b are zeroes of the polynomial

To find:

The value of k

Solution:

${x}^{2} - 7x + k$

We know that,

$Sum \: of \: zeroes = \frac{ - ( coefficient \: of \: x)}{coefficient \: of \: {x}^{2} } \\ \\ = > \: a \: + b \: = \frac{ -( - 7) }{1} \\ \\ = > a + b = 7 \: \: \: - - - - (eq1)</p><p>$

$Product \: of \: zeroes = \frac{constant \: term}{coefficient \: of \: {x}^{2} } \\ \\ = > ab \: = \frac{k}{1} \\ \\ = > ab \: = k$

We have

$a - b = 1 \: \: (given) - - - - (eq2)$

Now we will solve eq (1) and (2) by elimination method.

$a + b = 7 \\ a - b = 1 \: ( \: b \: cancelled \: ) \\we \: get \\2a = 8 \\ = > a = \frac{8}{2} \\ \\ = > a ={ 4}$

Substituting value of a in eq(1)

$a + b = 7 \\ 4 + b = 7 \\ = > b = 7 - 4 \\ \\ = > b = {3}$

Finding value of k

Putting values of a and b

We have ab = k (above solved)

So,

$ab = k \\ \\ 4 \times 3 = k \\ \\ = > k = 12$

Hence, the value of k is 12.

Hope it helps you

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