Page 61 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

√ √
2 2
2.35 The real numbers x, y satisfy the equation x − 2xy + y − 2x − 2y +6 = 0. Find
the minimum value of x + y .

Solution: Let x + y = k , then y = k − x . Substitute it into the equation:
√ √ √
x −2x(k−x)+(k−x) − 2x− 2(k−x)+6 = 0 ⇔ 4x −4kx+(k − 2k+6) = 0,
2
2
2
2
√ √ √
2 2
then Δ=(4k) − 16(k − 2k + 6) = 16 2k − 96 ≥ 0 ⇔ k ≥ 3 2,
√
thus k = x + y has the minimum value 3 2.
√ √
x
x
2.36 Solve the equation ( 2+ 3) +( 2 − 3) =4.
√ √ 360°
x
x
Solution: The equation is equivalent to ( 2+ 3) + √ 1 √ =4. Let y =( 2+ 3) ,
( 2+ 3) x
√
2
then y + 1 y =4 ⇒ y − 4y +1 = 0 whose roots are y =2 ± 3. thinking.
√ √ √ √
x
2
When y = 2+ 3, ( 2+ 3) = 2+ 3= ( 2+ 3) , thus x =2.
√
√ √ x 360° √ −2
1
√ =( 2+
thinking.
When y =2 − 3, ( 2+ 3) =2 − 3= 2+ 3 3) , thus x = −2.
We can verify that x =2,x = −2 are indeed roots of the original equation.

360°
thinking.

360°
thinking.

Discover the truth at www.deloitte.ca/careers Discover the truth at www.deloitte.ca/careers

© Deloitte & Touche LLP and affiliated entities.
Discover the truth at www.deloitte.ca/careers © Deloitte & Touche LLP and affiliated entities.

© Deloitte & Touche LLP and affiliated entities.
Discover the truth at www.deloitte.ca/careers
Download free eBooks at bookboon.com Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
61
61
© Deloitte & Touche LLP and affiliated entities.