Question

find the zeroes of the of the equation 4x square+ 24x+36 and verify the relationship between the zeroes and cofficient.​

Answer 1

Answer:

Given :-

• 4x² + 24x + 36

To Find :-

• What is the zeroes of a quadratic polynomial and verify the relationship between the zeroes and co-efficient.

Solution :-

Given Equation :

$\bigstar\: \: \bf{4x^2 + 24x + 36}$

Let,

$\implies \sf p(x) =\: 4x^2 + 24x + 36$

Zero of the polynomial is the value of x, where p(x) = 0

By putting p(x) = 0 we get,

$\implies \sf 4x^2 + 24x + 36 =\: 0$

$\implies \sf 4x^2 + (12 + 12)x + 36 =\: 0$

$\implies \sf 4x^2 + 12x + 12x + 36 =\: 0\: \: \bigg\lgroup \small\sf\bold{\pink{By\: splitting\: middle\: term}}\bigg\rgroup$

$\implies \sf 4x(x + 3) + 12(x + 3) =\: 0$

$\implies \sf (x + 3)(4x + 12) =\: 0$

$\longrightarrow \bf x + 3 =\: 0$

$\longrightarrow \sf\bold{\red{x =\: - 3}}$

$\longrightarrow \bf 4x + 12 =\: 0$

$\longrightarrow \sf 4x =\: - 12$

$\longrightarrow \sf x =\: \dfrac{- \cancel{12}}{\cancel{4}}$

$\longrightarrow \sf\bold{\red{x =\: - 3}}$

${\footnotesize{\bold{\underline{\therefore\: The\: zeroes\: of\: quadratic\: polynomial\: are\: - 3\: and\: - 3\: respectively\: .}}}}$

Hence,

• α = - 3
• β = - 3

Now, we have to verify the relationship between the zeroes and co-efficient :

Given Equation :

$\bigstar\: \: \bf 4x^2 + 24x + 36$

By comparing with ax² + bx + c we get,

• a = 4
• b = 24
• c = 36

❒ Sum Of Zeroes :

As we know that :

$\clubsuit$ Sum Of Zeroes Formula :

$\mapsto \sf\boxed{\bold{\pink{Sum\: Of\: Zeroes\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}$

We have :

• α = - 3
• β = - 3
• a = 4
• b = 24

According to the question by using the formula we get,

$\leadsto \sf (- 3) + (- 3) =\: \dfrac{- \cancel{24}}{\cancel{4}}$

$\leadsto \sf - 3 - 3 =\: \dfrac{- 6}{1}$

$\leadsto \sf\bold{\purple{- 6 =\: - 6}}$

Hence, Verified.

❒ Product Of Zeroes :

As we know that :

$\clubsuit$ Product Of Zeroes Formula :

$\mapsto \sf\boxed{\bold{\pink{Product\: Of\: Zeroes\: (\alpha\beta) =\: \dfrac{c}{a}}}}$

We have :

• α = - 3
• β = - 3
• a = 4
• c = 36

According to the question by using the formula we get,

$\leadsto \sf (- 3) \times (- 3) =\: \dfrac{\cancel{36}}{\cancel{4}}$

$\leadsto \sf\bold{\purple{9 =\: 9}}$

Hence, Verified.

Answer 2

$\underline {\underline{\bold{\red{ \star \: \: GIVEN :- }}}}$

$\footnotesize \bold { \mapsto \: 4x ^{2} + 24x + 36}$

$\underline {\underline{\bold{\pink{ \star\:TO\: FIND :- }}}}$

$\footnotesize\bold{ \mapsto \: What \: is \: the \: zeroes \: of \: a \: quadratic \: polynomial \: } \\ \footnotesize\bold{ \: \: \: \: \: \: \: and \: verify \: the \: relationship \: between \:the } \\ \footnotesize\bold{ \: \: \: \: \: \: \: zeroes \: and \: co-efficient. \: }\\ \\</p><p>$

$\underline {\underline{\bold{\orange{ \star \: \: SOLUTION :- }}}}$

$\footnotesize \bold \blue{P(x) = 4x ^{2} + 24x +36} \\ \\</p><p></p><p> \footnotesize \bold{Here, a = 4, b = 24, c = 36} \\ \\</p><p></p><p> \footnotesize \bold{p(x) = 4x ^{2} + 12x + 12x + 36} \\ \\</p><p></p><p> \footnotesize \bold{\Rightarrow \: 4x(x + 3) + 12(x + 3)} \\ \\</p><p></p><p> \footnotesize \bold{\Rightarrow \: (x+3)(4x+12)} \\ \\</p><p></p><p> \footnotesize \bold\red{\Rightarrow \: 4(x + 3)^{2} }$

$\footnotesize \bold{So, \: the \: zeros \: of \: quadratic \: polynomial \: is} \\ \footnotesize \bold{-3 \: and \: -3.} \\ \\</p><p> \: \footnotesize \bold{ Let \: \alpha \: = -3 \: and \: \beta = - 3} \\ \\</p><p></p><p>\footnotesize \bold{Sum \: of \: zeros = \alpha \: + \: \beta = - 3 - 3 = - 6 } \\ \\\footnotesize \bold{ Sum of zeros = \frac{coefficient \: of \: x}{ coefficient \: of \: {x}^{2} }} \\ \\ \footnotesize \bold{ - \frac{b}{a} = \frac{ - 24}{4} = - 6} \\ \\\footnotesize \bold \red{So, sum \: of \: zeros = a + B = \frac{coefficient \: of \: x}{ coefficient \: of \: {x}^{2} } \: \: }$

$\footnotesize \bold{Product \: of \: zeros= \alpha \beta = (-3)(-3) = 9} \\ \\\footnotesize \bold{Also, product \: of \: zeros = \frac{constant \: term}{coefficient of {x}^{2} } }\\ \footnotesize \bold{ = \frac{c}{a} = \frac{36}{4} = 9}\\ \\\footnotesize \bold \red{so \: product \: of \: zeros = \alpha \beta = \frac{constant \: term}{coefficient of {x}^{2} } }$