Question

If the squared difference of the zeros of the quadratic polynomial f(x)= x²+px+45 is equal to 144, find the value of p.​

Answer 1

❍ Let's Consider $\bf \alpha$ and $\bf \beta$ be the two zeroes of Polynomial f(x)= x²+ px + 45 .

⠀⠀⠀⠀⠀P O L Y N O M I A L : f(x)= x²+ px + 45

Here ,

• Cofficient of x² : 1
• Cofficient of x : p
• Constant term : 45

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \maltese \:\: \bf Sum \:of \:zeroes\::$

$\qquad \dag\:\: \bigg\lgroup \sf{ \alpha + \beta = \dfrac{ -( Cofficient\: of\: x^2)}{ Cofficient \:of\:x}}\bigg\rgroup$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf \alpha + \beta = \dfrac{ -( Cofficient\: of\: x)}{ Cofficient \:of\:x^2}$

$\qquad:\implies \sf \alpha + \beta = \dfrac{ -( p)}{1 }$

$\qquad:\implies \sf \alpha + \beta = \dfrac{ - p}{1 }$

$\qquad:\implies \bf \alpha + \beta = -p \qquad \bigg\lgroup \sf{ \:Equation \:1 }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀&

$\dag\:\:\sf{ As,\:We\:know\:that\::} \\\\ \qquad \maltese\:\: \bf Product \:of \:zeroes\::$

$\qquad \dag\:\: \bigg\lgroup \sf{ \alpha \beta = \dfrac{Constant\:Term \:}{ Cofficient \:of\:x^2}}\bigg\rgroup$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf \alpha \beta = \dfrac{Constant\:Term \:}{ Cofficient \:of\:x^2}$

$\qquad:\implies \sf \alpha \beta = \dfrac{45\:}{ 1}$

$\qquad:\implies \bf \alpha \beta = 45 \qquad \bigg\lgroup \sf{ \:Equation \:2 }\bigg\rgroup$

Now ,

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:According \: to\: \: the \: Question \::}}$

⠀⠀⠀⠀⠀━━ If the squared difference of the zeros of the quadratic polynomial is equal to 144.

$\qquad:\implies \sf ( \alpha - \beta )^2 = 144$

$\qquad:\implies \sf ( \alpha + \beta )^2 - 4 \alpha \beta = 144$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Eq.1 \:\:\& \:\:Eq.2 \: \::}}$

$\qquad:\implies \sf ( \alpha + \beta )^2 - 4 \alpha \beta = 144$

$\qquad:\implies \sf (-p)^2 - 4 \alpha \beta = 144$

$\qquad:\implies \sf (-p)^2 - 4 \times 45 = 144$

$\qquad:\implies \sf p^2 - 180= 144$

$\qquad:\implies \sf p^2 = 144 + 180$

$\qquad:\implies \sf p^2 = 324$

$\qquad:\implies \sf p = \sqrt{324}$

$\qquad:\implies \sf p = \pm 18$

$\qquad :\implies \frak{\underline{\purple{\:p = \pm 18\:\: }} }\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The\:Value\:of\:p \:is\:\bf{ \pm 18}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Answer 2

According to the question we know that The given quadratic polynomial is fx = x²+px+45

we can say that α+ β= -p

and also, say αβ = 45

Now :-

we can show that

( α + β )² - 4αβ = (α - β )²

(α + β)² - 4αβ = 144

we know that (α+β)² = -p and αβ = 44 then we put it value in given equation

So :-

p² - 4 × 45 = 144

p ² - 180 = 144

p² = 144+ 180

p² = 324

p = √324

p = ±18

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