Question

If one root is equal to the square of other root of the equation x² + x - k = 0, then k is
a) 0
b) 1
c)-1
d) 2
​

Answer 1

Answer:

k=-1

Step-by-step explanation:

let one root be a , other root is a^2

sum of the roots = a+ a^2 = - b/a = -1

so a^2 = -1- a

product of the roots = a^3= -k or a(-1-a) = -k

i.e. ,-a-a^2 = -k or -a-(-1-a) = -k or k = -1

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hope it helps

well it can be wrong

Answer 2

GIVEN :–

• One root is equal to the square of other root of the equation x² + x - k = 0.

TO FIND :–

• Value of 'k' = ?

SOLUTION :–

• We know that –

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

$\\ \bf \implies \alpha + { \alpha }^{2} = \dfrac{ - (1)}{1}$

$\\ \bf \implies \alpha + { \alpha }^{2} = - 1$

$\\ \bf \implies { \alpha }^{2} + \alpha + 1 = 0$

$\\ \bf \implies \alpha = \dfrac{ - 1 \pm \sqrt{ {(1)}^{2} - 4(1)(1)} }{2(1)}$

$\\ \bf \implies \alpha = \dfrac{ - 1 \pm \sqrt{1-4}}{2}$

$\\ \bf \implies \alpha = \dfrac{ - 1 \pm \sqrt{ - 3}}{2}$

$\\ \bf \implies \alpha = \dfrac{ - 1 \pm \sqrt{3}i}{2}$

$\\ \bf \implies \alpha = \dfrac{ - 1 + \sqrt{3}i}{2},\dfrac{ - 1 - \sqrt{3}i}{2}$

• We also know that –

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

$\\ \bf \implies ( \alpha )( { \alpha }^{2})= \dfrac{ - k}{1}$

$\\ \bf \implies k = - { \alpha }^{3}$

▪︎When $\bf\alpha = \dfrac{ - 1 + \sqrt{3}i}{2}$ :–

• We should write this as –

$\\ \bf \implies k = - \bigg[\dfrac{ - 1 + \sqrt{3}i}{2} \bigg]\bigg[\dfrac{ - 1 + \sqrt{3}i}{2} \bigg] ^{2}$

$\\ \bf \implies k = - \bigg[\dfrac{ - 1 + \sqrt{3}i}{2} \bigg]\bigg[\dfrac{ - 1 - \sqrt{3}i}{2} \bigg]$

$\\ \bf \implies k = - \dfrac{1}{4} ( - 1 + \sqrt{3}i)(- 1 - \sqrt{3}i)$

$\\ \bf \implies k = - \dfrac{1}{4} \{ ( - 1)^{2} - (\sqrt{3}i)^{2} \}$

$\\ \bf \implies k = - \dfrac{1}{4} \{1- 3(i)^{2} \}$

$\\\bf \implies k = - \dfrac{1}{4} \{1- 3( - 1)\}$

$\\ \bf \implies k = - \dfrac{1}{4} (1 + 3)$

$\\ \bf \implies k = - \dfrac{1}{4} (4)$

$\\ \large\implies{ \boxed{ \bf k = -1}}$

▪︎When $\bf\alpha = \dfrac{ - 1 -\sqrt{3}i}{2}$ :–

• We should write this as –

$\\ \bf \implies k = - \bigg[\dfrac{ - 1- \sqrt{3}i}{2} \bigg]\bigg[\dfrac{ - 1 -\sqrt{3}i}{2} \bigg] ^{2}$

$\\ \bf \implies k = - \bigg[\dfrac{ - 1 - \sqrt{3}i}{2} \bigg]\bigg[\dfrac{ - 1+ \sqrt{3}i}{2} \bigg]$

$\\ \bf \implies k = - \dfrac{1}{4} ( - 1 - \sqrt{3}i)(- 1 + \sqrt{3}i)$

$\\ \bf \implies k = - \dfrac{1}{4} \{ ( - 1)^{2} - (\sqrt{3}i)^{2} \}$

$\\ \bf \implies k = - \dfrac{1}{4} \{1- 3(i)^{2} \}$

$\\ \bf \implies k = - \dfrac{1}{4} \{1- 3( - 1)\}$

$\\ \bf \implies k = - \dfrac{1}{4} (1 + 3)$

$\\ \bf \implies k = - \dfrac{1}{4} (4)$

$\\ \large\implies{ \boxed{ \bf k = -1}}$

• Hence Option (c) is correct.