Question

If the square of difference of the zeros of the quadratic polynomial x2+px+45 is equal to 144 then find p

Answer 1

Answer:

+18

Step-by-step explanation:

Let α,β be the zeroes of the given quadratic polynomial f(x)=x  

2

+px+45

∴α+β=−p,αβ=45

It is given that, (α−β)  

2

=144

⇒(α+β)  

2

–4αβ=144

⇒(–p)  

2

–4×45=144

⇒p  

2

–180=144

⇒p  

2

=144+180

⇒p  

2

=324

⇒p=±  

324

​

 

∴p=±18

Answer 2

Given

• Square of difference of the zeros of the quadratic polynomial x² + px + 45 is 144.

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To Find

• The value of 'p'.

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Solution

In the given polynomial, we are aware of the fact that -:

α + β = -p

αβ = 45

So, according to the question,

(α - β)² = 144    -(i)

Apply the below formula to find the answer.

(α - β)² = (α + β)² - 4αβ

Substitute the value of α + β and αβ to continue further.

⇒ (α - β)² = (p)² - 4(45)

⇒ (α - β)² = p² - 180    - (ii)

Taking (i) and (ii)

p² - 180 = 144

p² = 144 + 180

p² = 324

p = √324

p = ±18

∴ The value of 'p' can be 18 or -18.

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