Question

Find the value of p, for which one zero of the polynomial, px - 144x + 8 is 11 times the other.
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Answer 1

Question :

Find the value of p, for which one zero of the polynomial, px² - 144x + 8 is 11 times the other.  

Given :

• polynomial => px² - 144x + 8
• One zero is 11 times the other zero { zero of polynomial }  

To find :

Value of p  

Method and formula used : 

For general quadratic equations represented by ax² + bx + c = 0 , If α and β are zeros of above mentioned equation .

Then,

• Products of zero (αβ) = c/a
• Sum of zeros (α + β) = (-b/a)  

Solution :  

Let α and β be zeros of px² - 144x + 8  such that α = 11× β ........ equation 1  Also , on comparing px² - 144x + 8 = 0  with General quadratic equation we get;

a = p

b = -144

c = 8

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Sum of zeros = -b/a

α + β = - { (-144)/p }

Putting value of α from Equation 1 ,into above equation we get;

➝ (11×β) + β = 144/p

➝ 12β = 144/p

➝ β × p = 144 ÷ 12

➝ β × p = 12......... equation 2

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Product of zeros = c/a

➝ α × β  = 8/p

➝  α × β×p  = 8

On Putting value of (β×p)  from Equation 2 , into above equation we get;

➝ α × (12) = 8

➝ α = 8/12

➝ α = 2/3

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On putting value of α In equation 1 , we get;

➝ (2/3) = 11× β

➝ β = 2 ÷(3×11)  ➝ β = 2/33 _______________________________________________

From Equation 2 we know that ,β × p = 12

➝ p = 12/β

➝ p = 12 ÷ (2/33)

➝ p = 12 × 33/2

➝ p = 6 × 33

➝ p = 198

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ANSWER :

p = 198