X 2-29 x -210y 2-28 y -195A- eB- aC- bD- cE- d

The correct option is E (d) x^2 + 29x = 210 x^2 + 29x + 210 = 0 x^2+ 15x + 14x + 210 = 0 [x + 15][x + 14] = 0 x = 15, 14 y^2 + 28y = 195 y^2 + 28y + 195 = ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

x 2+29 x =-21...

Question

$x^{2} + 29 x = - 210$

$y^{2} + 28 y = - 195$

(a)

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(b)

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(c)

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(d)

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(e)

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Solution

The correct option is D
(d)

$x^{2}$ + 29x = -210

$x^{2}$ + 29x + 210 = 0

$x^{2}$ + 15x + 14x + 210 = 0

[x + 15][x + 14] = 0

x = -15, -14

$y^{2}$ + 28y = -195

$y^{2}$ + 28y + 195 = 0

$y^{2}$ + 15y + 13y + 195 = 0

[y + 15][y + 13] = 0

y = -15, -13

x = -15, -14

So, clearly, $x \leq y$

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Similar questions

$x^{2}$ + 29x = -210

$y^{2}$ + 28y = -195

Q. In ∆ABC, BD ⊥ AC and CE ⊥ AB such that BE = CD. Prove that BD = CE.

△ A B C

is iscoeles in which

A B = A C

, seg

B D

and seg

C E

are medians show that

B D = C E

Q. ABC is an isosceles triangle with AB=AC AND BD,CE are it's two median . Show that BD=CE.

∆

ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.

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x2+29x=−210 y2+28y=−195

The correct option is D (d) x2 + 29x = -210 x2 + 29x + 210 = 0 x2+ 15x + 14x + 210 = 0 [x + 15][x + 14] = 0 x = -15, -14 y2 + 28y = -195 y2 + 28y + 195 = 0 y2 + 15y + 13y + 195 = 0 [y + 15][y + 13] = 0 y = -15, -13 x = -15, -14 So, clearly, x≤y

$x^{2} + 29 x = - 210$

$y^{2} + 28 y = - 195$

The correct option is D
(d)

$x^{2}$ + 29x = -210

$x^{2}$ + 29x + 210 = 0

$x^{2}$ + 15x + 14x + 210 = 0

[x + 15][x + 14] = 0

x = -15, -14

$y^{2}$ + 28y = -195

$y^{2}$ + 28y + 195 = 0

$y^{2}$ + 15y + 13y + 195 = 0

[y + 15][y + 13] = 0

y = -15, -13

x = -15, -14

So, clearly, $x \leq y$