Question

Find the value of P for which one root of the quadratic equation Px²- 14x+8=0 is 6 times the other.

Answer 1

Question Given:

Px² - 14x + 8=0

Also, let α be a root of the above equation 


Solution:

As It is given one root is 6 times the other and so the roots would be

α and 6α

Let us take sum of zeroes/roots for the equation Px² - 14x + 8 = 0

Then,   α + 6α = -b/a

                  7α = -(-14)/P
 
                    α = 2/P                            ..............(1)

Now taking product of zeroes for the equation Px² - 14x + 8=0

Then,   α × 6α = c/a

                 6α² = 8/P                                .............(2)

From (1) and (2)

           6(2/P)² = 8/P

         6 × 4/P² = 8/P
 
             24/P² = 8/P
     
                24P = 8P²

(÷8) ⇒      P² = 3P

= P² - 3P

= P² - 3P + 0P + 0

= P(P-3) + 0(P-3)

= (P+0)(P-3)

Then, P = 3, 0

If P is 0 it cannot form a quadratic equation and goes like this 

=0x²- 14x + 8=0
   
= -14x + 8        Which is not a quadratic Equation 

So, Neglect 0, Then P = 3


Answer : P =3

☺ Hope this Helps ☺

Answer 2

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