Question

If a polynomial
p
(
x
)
=
x
2
−
a
x
+
b
has two zeroes 1 and 7, what is the value of
a
b
?

Answer 1

Answer:

a=8 , b=7.

Step-by-step explanation:

Given,

p(x) = x²-ax+b

Now consider,

p(1) = (1)²-a(1)+b

= 1-a+b

-a+b+1 = 0 → (1)

Now consider,

p(7) = (7)²-a(7)+b

= 49-7a+b

-7a+b+49 = 0 → (2)

By solving equation (1)&(2)

-a + b + 1 = 0

-7a + b + 49 = 0

(+) (-) (-)

_____________

6a - 48 = 0

6a = 48

a = 48/6

:. a = 8

Substitute 'a' value in equation (1)

-a + b + 1 = 0

-(8)+b+1 = 0

-8+b+1 = 0

-7 + b = 0

:. b = 7

:. a = 8,b = 7.

I hope it helps you.

Answer 2

Corrected Question :

• If a polynomial p(x) = x^2 - ax + b has two zeros 1 and 7, what is the value of a and b.

Answer :

The quadratic polynomial given is p(x) = x² - ax + b

The zeros of the polynomial is 1 and 7.

Let alpha = 1 and beta = 7.

According to the relationship between the zeros and the coefficients of a quadratic polynomial :-

→ α + β = 1 + 7 = 8

→ α × β = 1 × 7 = 7

Now, the coefficients of the given quadratic equation will be :

→ x² - [Sum of zeros] x + Product of zeros

→ x² - 8x + 7

Therefore the values of a and b are -8 and 7 respectively.

Verification :

We know that :

→ α + β = - b/a

→ α × β = c/a

To verify :

→ α + β = 1 + 7 = 8 = - (-8)/1 = -b/a

→ α × β = 1 x 7 = 7/1 = c/a

So, it is verified that the answer is correct.