Question

$\dag \: \: \small \rm \sqrt{13} + \sqrt{ {k}^{2} - 2k + 5} = \sqrt{ {k}^{2} + 4k + 20}$
For the above attached equation, find the value of - 'k'.



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Answer 1

ANSWER:

• The value of k = 4

GIVEN:

$\sf \sqrt{13} + \sqrt{ {k}^{2} - 2k + 5} = \sqrt{ {k}^{2} + 4k + 20}$

TO FIND:

• The value of k.

EXPLANATION:

$\sf Let\ A = \sqrt{13} + \sqrt{ {k}^{2} - 2k + 5}$

$\sf Let\ B=\sqrt{ {k}^{2} + 4k + 20}$

$\boxed{\huge{\bold{\gray{A=B}}}}$

Squaring on both sides.

$\boxed{\huge{\bold{\gray{A^{2} =B ^{2} }}}}$

Take A:

$\sf A = \sqrt{13} + \sqrt{ {k}^{2} - 2k + 5}$

Squaring on both sides.

$\boxed{\bold{ \large{\gray{(X + Y)^2 = X^2 + 2XY + Y^2}}}}$

$\sf A^{2} = 13 + k^2 - 2k +5 +2\sqrt{13(k^2-2k+5}$

$\sf A^{2} = 18 + k^2 - 2k +2\sqrt{13k^2-26k+65}$

Take B:

$\sf B=\sqrt{ {k}^{2} + 4k + 20}$

Squaring on both sides.

$\sf B^{2} = {k}^{2} + 4k + 20$

Equate A² and B².

Final Equation:

4k² - 32k + 64 = 0

Divide by 4 on both sides.

k² - 8k + 16 = 0

k² - 2(4)(k) + 4² = 0

It is in the form of x² - 2xy + y²

(k - 4)² = 0

k - 4 = 0

k = 4.

HENCE THE VALUE OF K = 4

VERIFICATION:

Substitute k = 4 in given question.

$\sf \sqrt{13} + \sqrt{ {4}^{2} - 2(4) + 5} = \sqrt{ {4}^{2} + 4(4) + 20}$

$\sf \sqrt{13} + \sqrt{ 16- 8 + 5} = \sqrt{ 16 + 16+ 20}$

$\sf \sqrt{13} + \sqrt{ 8 + 5} = \sqrt{52}$

$\sf \sqrt{13} + \sqrt{13} = \sqrt{4\times 13}$

$\sf 2\ \sqrt{13}= 2\ \sqrt{13}$

Hence verified.

NOTE : REFER ATTACHMENT FOR DETAILED SOLUTION.

Answer 2

Answer:

Since the above equation has equal roots

Hence

B2−4AC=0

Hence

4(2k−3) 2−4(5k−6)(k−2)=0

4k 2 −12k+9−(5k 2 −16k+12)=0

−k 2+4k−3=0

OR

k2 −4k+3=0

(k−3)(k−1)=0

Hence

k=3 and k=1.

Step-by-step explanation:

Hope it will help you..