Math, asked by tomarramesh957, 1 year ago

If a and b are zeros of polynomial px2-5x+q ,then value of p and q, id a+b=10​

Answers

Answered by nickcolldagar20
53

Answer:

We, are provided with the following information:

a + b = 10

px ^{2}  - 5x + q = 0

Now, we know that

 \alpha  +  \beta  =  -  \frac{b}{a}

Therefore:

10 =  -  \frac{ - 5}{p}

p =  \frac{1}{2}

Answered by slicergiza
17

Value of p is 1/2 and value of q can't be determined

Step-by-step explanation:

Given,

a and b are zeros of polynomial px²-5x+q,

Then,

\text{Sum of roots}=-\frac{\text{Coefficient of x}}{\text{Coefficient of }x^2}

a+b=-\frac{-5}{p}

a+b=\frac{5}{p}

We have,

a + b = 10,     ...(1)

10 = \frac{5}{p}

\implies p = \frac{5}{10}=\frac{1}{2}

Now,

if a and b are roots of the

\text{Product of roots} =\frac{\text{Constant}}{\text{Coefficient of }x^2}

ab=\frac{q}{p}    

ab =2q         ....(2)

∵ a is a root  of px^2 - 5x + q=0

\implies pa^2 - 5a + q = 0

 \frac{1}{2}a^2 - 5a + \frac{ab}{2}=0

a^2 - 10a + ab = 0  ...(3)

Similarly b is a root of px^2 - 5x + q

\implies b^2 - 10b + ab =0 ....(4)

Eq (3) - Eq (4)

a^2 - b^2 - 10 a + 10b =0

(a^2-b^2) - 10(a-b)=0

(a+b)(a-b) - 10(a-b)=0

(a-b)(a+b-10)=0

(a-b)(10-10)=0

(a-b)(0)=0

Thus, a -b can not be find.

That is, value of q can't find with the given information.

#Learn more:

If the sum of zeroes of a polynomial px2+5x+8p is equal to the product of zeroes find the value of p.

https://brainly.in/question/3064620

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