Question

If the squared difference of the zeroes of the quadratic polynomial f(x)=x^2+px+45 is equal to 144, find the value of p​

Answer 1

Answer:

Step-by-step explanation:

Given :-

Quadratic Polynomial fx = x² + px + 45

To Find :-

Value of p.

Solution :-

We can write α + β = - p ... (i)

And, αβ = 45 ... (ii)

We can show that,

(α + β)² - 4αβ = (α - β)²

(α + β)² - 4αβ = 144

Now, putting Eq (i) and (ii), we get

⇒ p² - 4 × 45 = 144

⇒ p ² - 180 = 144

⇒  p² = 144 + 180

⇒ p² = 324

⇒ p = √324

⇒ p = ± 18

Hence, the value of p​ is 18.

Answer 2

Answer:

• Let the $\alpha$ and $\beta$ be the Zeroes of the Quadratic Polynomial.
• $\sf (\alpha-\beta)^2=144$

$\underline{\bigstar\:\:\textsf{By the Use of Formula :}}$

$\dashrightarrow\tt\:\:(\alpha + \beta)^2 - (\alpha - \beta)^2 = 4\alpha \beta\\\\\\\dashrightarrow\tt\:\:\bigg(\dfrac{- \:b}{a}\bigg)^2 - (\alpha - \beta)^2= 4 \times \dfrac{c}{a} \\\\ {\scriptsize\qquad\bf{\dag}\:\:\texttt{Here, a = 1 , b = p , c = 45}}\\\\\\\dashrightarrow\tt\:\:\bigg(\dfrac{- \:p}{1}\bigg)^2 - 144= 4 \times \dfrac{45}{1}\\\\\\\dashrightarrow\tt\:\:( - p)^2 - 144 = 180\\\\\\\dashrightarrow\tt\:\:p^2 = 180 + 144\\\\\\\dashrightarrow\tt\:\:p^2 = 324\\\\\\\dashrightarrow\tt\:\:p = \sqrt{324}\\\\\\\dashrightarrow\:\:\underline{\boxed{\tt p = \pm\:18}}$

⠀

$\therefore\:\underline{\textsf{Therefore, value of p is \textbf{ \pm 18}}}.$