Question

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Answer 1

GIVEN :–

• A quadratic equation ax²+bx+c = 0 (a ≠0) .

• One root is 3 times of other root.

TO FIND :–

• Value of 16ac = ?

SOLUTION :–

• Let the root of quadratic equation = α

• Other root of quadratic equation = 3α

▪︎ We know that –

$\\\implies\bf Sum \: \: of \: \: roots = - \dfrac{Coefficient \: \: of \: \: x}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf \alpha + 3 \alpha = - \dfrac{b}{a}$

$\\\implies\bf 4 \alpha = - \dfrac{b}{a}$

$\\\implies\bf \alpha = - \dfrac{b}{4a} \: \: \: \: \: \: - - - - eq.(1)$

▪︎ We also know that –

$\\\implies\bf Product \: \: of \: \: roots = \dfrac{Constant \: \: term}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf (\alpha )(3 \alpha )= \dfrac{c}{a}$

$\\\implies\bf 3{\alpha}^{2} = \dfrac{c}{a}$

$\\\implies\bf {\alpha}^{2} = \dfrac{c}{3a}$

• Using eq.(1) –

$\\\implies\bf { \bigg(-\dfrac{b}{4a} \bigg)}^{2} = \dfrac{c}{3a}$

$\\\implies\bf \dfrac{ {b}^{2} }{16 {a}^{2} } = \dfrac{c}{3a}$

$\\\implies\bf3a{b}^{2} = 16 {a}^{2}c$

$\\\implies\bf3{b}^{2} = 16ac$

• Hence –

$\\\implies \large { \boxed{\bf16ac = 3{b}^{2}}}$

Answer 2

Answer:

SOLUTION :–

• Let the root of quadratic equation = α

• Other root of quadratic equation = 3α

▪︎ We know that –

$\\\implies\bf Sum \: \: of \: \: roots = - \dfrac{Coefficient \: \: of \: \: x}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf \alpha + 3 \alpha = - \dfrac{b}{a}$

$\\\implies\bf 4 \alpha = - \dfrac{b}{a}$

$\\\implies\bf \alpha = - \dfrac{b}{4a} \: \: \: \: \: \: - - - - eq.(1)$

▪︎ We also know that –

$\\\implies\bf Product \: \: of \: \: roots = \dfrac{Constant \: \: term}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf (\alpha )(3 \alpha )= \dfrac{c}{a}$

$\\\implies\bf 3{\alpha}^{2} = \dfrac{c}{a}$

$\\\implies\bf {\alpha}^{2} = \dfrac{c}{3a}$

• Using eq.(1) –

$\\\implies\bf { \bigg(-\dfrac{b}{4a} \bigg)}^{2} = \dfrac{c}{3a}$

$\\\implies\bf \dfrac{ {b}^{2} }{16 {a}^{2} } = \dfrac{c}{3a}$

$\\\implies\bf3a{b}^{2} = 16 {a}^{2}c$

$\\\implies\bf3{b}^{2} = 16ac$

• Hence –

$\\\implies \large { \boxed{\bf16ac = 3{b}^{2}}}$

Answer 3

Answer:

SOLUTION :–

• Let the root of quadratic equation = α

• Other root of quadratic equation = 3α

▪︎ We know that –

$\\\implies\bf Sum \: \: of \: \: roots = - \dfrac{Coefficient \: \: of \: \: x}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf \alpha + 3 \alpha = - \dfrac{b}{a}$

$\\\implies\bf 4 \alpha = - \dfrac{b}{a}$

$\\\implies\bf \alpha = - \dfrac{b}{4a} \: \: \: \: \: \: - - - - eq.(1)$

▪︎ We also know that –

$\\\implies\bf Product \: \: of \: \: roots = \dfrac{Constant \: \: term}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf (\alpha )(3 \alpha )= \dfrac{c}{a}$

$\\\implies\bf 3{\alpha}^{2} = \dfrac{c}{a}$

$\\\implies\bf {\alpha}^{2} = \dfrac{c}{3a}$

• Using eq.(1) –

$\\\implies\bf { \bigg(-\dfrac{b}{4a} \bigg)}^{2} = \dfrac{c}{3a}$

$\\\implies\bf \dfrac{ {b}^{2} }{16 {a}^{2} } = \dfrac{c}{3a}$

$\\\implies\bf3a{b}^{2} = 16 {a}^{2}c$

$\\\implies\bf3{b}^{2} = 16ac$

• Hence –

$\\\implies \large { \boxed{\bf16ac = 3{b}^{2}}}$

Answer 4

Answer:

SOLUTION :–

• Let the root of quadratic equation = α

• Other root of quadratic equation = 3α

▪︎ We know that –

$\\\implies\bf Sum \: \: of \: \: roots = - \dfrac{Coefficient \: \: of \: \: x}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf \alpha + 3 \alpha = - \dfrac{b}{a}$

$\\\implies\bf 4 \alpha = - \dfrac{b}{a}$

$\\\implies\bf \alpha = - \dfrac{b}{4a} \: \: \: \: \: \: - - - - eq.(1)$

▪︎ We also know that –

$\\\implies\bf Product \: \: of \: \: roots = \dfrac{Constant \: \: term}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf (\alpha )(3 \alpha )= \dfrac{c}{a}$

$\\\implies\bf 3{\alpha}^{2} = \dfrac{c}{a}$

$\\\implies\bf {\alpha}^{2} = \dfrac{c}{3a}$

• Using eq.(1) –

$\\\implies\bf { \bigg(-\dfrac{b}{4a} \bigg)}^{2} = \dfrac{c}{3a}$

$\\\implies\bf \dfrac{ {b}^{2} }{16 {a}^{2} } = \dfrac{c}{3a}$

$\\\implies\bf3a{b}^{2} = 16 {a}^{2}c$

$\\\implies\bf3{b}^{2} = 16ac$

• Hence –

$\\\implies \large { \boxed{\bf16ac = 3{b}^{2}}}$

Answer 5

Answer:

SOLUTION :–

• Let the root of quadratic equation = α

• Other root of quadratic equation = 3α

▪︎ We know that –

$\\\implies\bf Sum \: \: of \: \: roots = - \dfrac{Coefficient \: \: of \: \: x}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf \alpha + 3 \alpha = - \dfrac{b}{a}$

$\\\implies\bf 4 \alpha = - \dfrac{b}{a}$

$\\\implies\bf \alpha = - \dfrac{b}{4a} \: \: \: \: \: \: - - - - eq.(1)$

▪︎ We also know that –

$\\\implies\bf Product \: \: of \: \: roots = \dfrac{Constant \: \: term}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf (\alpha )(3 \alpha )= \dfrac{c}{a}$

$\\\implies\bf 3{\alpha}^{2} = \dfrac{c}{a}$

$\\\implies\bf {\alpha}^{2} = \dfrac{c}{3a}$

• Using eq.(1) –

$\\\implies\bf { \bigg(-\dfrac{b}{4a} \bigg)}^{2} = \dfrac{c}{3a}$

$\\\implies\bf \dfrac{ {b}^{2} }{16 {a}^{2} } = \dfrac{c}{3a}$

$\\\implies\bf3a{b}^{2} = 16 {a}^{2}c$

$\\\implies\bf3{b}^{2} = 16ac$

• Hence –

$\\\implies \large { \boxed{\bf16ac = 3{b}^{2}}}$

Answer 6

Answer:

SOLUTION :–

• Let the root of quadratic equation = α

• Other root of quadratic equation = 3α

▪︎ We know that –

$\\\implies\bf Sum \: \: of \: \: roots = - \dfrac{Coefficient \: \: of \: \: x}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf \alpha + 3 \alpha = - \dfrac{b}{a}$

$\\\implies\bf 4 \alpha = - \dfrac{b}{a}$

$\\\implies\bf \alpha = - \dfrac{b}{4a} \: \: \: \: \: \: - - - - eq.(1)$

▪︎ We also know that –

$\\\implies\bf Product \: \: of \: \: roots = \dfrac{Constant \: \: term}{Coefficient \: \: of \: \: {x}^{2} }$

$\\\implies\bf (\alpha )(3 \alpha )= \dfrac{c}{a}$

$\\\implies\bf 3{\alpha}^{2} = \dfrac{c}{a}$

$\\\implies\bf {\alpha}^{2} = \dfrac{c}{3a}$

• Using eq.(1) –

$\\\implies\bf { \bigg(-\dfrac{b}{4a} \bigg)}^{2} = \dfrac{c}{3a}$

$\\\implies\bf \dfrac{ {b}^{2} }{16 {a}^{2} } = \dfrac{c}{3a}$

$\\\implies\bf3a{b}^{2} = 16 {a}^{2}c$

$\\\implies\bf3{b}^{2} = 16ac$

• Hence –

$\\\implies \large { \boxed{\bf16ac = 3{b}^{2}}}$