Question

If the zeros of the polynomial x² - px + q,
be in the ratio 2 : 3, prove that 6p² = 25q​

Answer 1

f(x) = x² - px + q and it's zeros are in the ratio 2 : 3.

__________ [ GIVEN ]

• We have to prove that 6p² = 25q

___________________________

• Let the zeros be 2M and 3M.

f(x) = x² - px + q

Here..

a = 1, b = -p, c = q

Now..

Sum of zeros = -b/a

→ 2M + 3M = -(-p)/1

→ 5M = p/1

→ 5M = p ________ (eq 1)

Product of zeros = c/a

→ 2M × 3M = q/1

→ 6M² = q

→ M² = q/6

→ M = √(q/6)

Put value of M in (eq 1)

→ 5[√(q/6)] = p

We can write it like also,,

→ 5(√q/√6) = p

Cross-multiply them

→ 5√q = p√6

Squaring on both sides

→ (5√q)² = (p√6)²

→ 25q = p²6

→ 25q = 6p²

___ [ HENCE PROVED ]

______________________________

Answer 2

$\huge\boxed{{{Solution:-}}}$

According to the given question:-

• 6p² = 25q have to prove
• Let the zeros be 2Y and 3Y
• f(x) = x² - px + q

So,

• a = 1, b = -p, c = q

Then,

-b/a = sum of zeros

=> 2Y + 3Y = -(-p)/1

=> 5Y = p/1

=> 5Y = p __Equation - (1)

c/a = product of zeros

=> 2Y × 3Y = q/1

=> 6Y² = q

=> Y² = q / 6

=> Y = √(q / 6)

Adding values of Y in _ equation - (1)

=> 5(√(q / 6) = p

=> 5(√q / √6) = p

Cross-multiply

=> 5 √ q = p √ 6

Squaring Both sides:-

=> (5√q)² = (p√6)²

=> 25q = p²6

=> 25q = 6p²

$\boxed{{{Therefore-Proved}}}$