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

Given:-

→ Polynomial : f(x) = x² + px + 45

→ Squared difference of zeros = 144

To find:-

→ Value of p?

Solution:-

Let the zeros be α & β

A/q,

→ (α - β)² = 144

→ α - β = √144

→ α - β = 144

Here, sum of zeros:-

→ α + β = -p/1

→ α + β = -p

Now, product of zeros:-

→ αβ = 45/1

→ αβ = 45

We know,

(a - b)² = (a + b)² - 4ab

Now putting all values:-

→ (12)² = (-p)² - 4 × 45

→ 144 = p² - 180

→ p² = 144 + 180

→ p² = 324

→ p = √324

→ p = ± 18

Therefore,

Required value of p = ± 18 .

Answer 2

Answer:

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

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

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

⠀

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