Question

find the value of p such that the quadratic equation x^2-(p+6)x+2(2p-1) =0 has sum of the roots as half of their product​

Answer 1

Answer:-

Given: equation is x² - (p + 6)x + 2(2p - 1) = 0.

Let ,

• a = 1

• b = - (p + 6)

• c = 2(2p - 1)

And,

Sum of the roots is half of their product.

We know that,

Sum of the zeroes = - b/a

→ sum of the zeroes = - [ - (p + 6) ] / 1

→ sum of the zeroes = p + 6

Product of the zeroes = c/a

→ Product of the zeroes = 2(2p - 1)

Hence,

→ p + 6 = 1/2 * 2(2p - 1)

→ p + 6 = 2p - 1

→ 6 + 1 = 2p - p

→ 7 = p

Therefore, the value of p is 7.

Answer 2

Answer:

• Equation : x² - (p + 6)x + 2(2p - 1) = 0
• a = 1
• b = - (p + 6)
• c = 2(2p - 1)

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf Sum \:of \:Roots = \dfrac{1}{2} \times Product \:of \:Roots\\\\\\:\implies\sf \dfrac{ -b}{a} = \dfrac{1}{2} \times \dfrac{c}{a}\\\\\\:\implies\sf - b = \dfrac{1}{2} \times c\\\\\\:\implies\sf -[-(p+6)]= \dfrac{1}{2} \times 2(2p-1)\\\\\\:\implies\sf p + 6 = 2p - 1\\\\\\:\implies\sf 6 + 1 = 2p - p\\\\\\:\implies\underline{\boxed{\sf p = 7}}$

⠀

$\therefore\:\underline{\textsf{Therefore, required value of p is \textbf{7}}}.$