All categories

3 years ago

Find the values of a and b such that x^2 + 3x + 4 = (x + a)^2+ b

Mathematics

1 answer:

sesenic [268]3 years ago

8 0

Answer:

$x {}^{2} + 3x + 4 = (x + a) {}^{2} + b \\ x {}^{2} + 3x + 4 = x {}^{2} + 2ax + a {}^{2} + b \\ = x {}^{2} + a {}^{2} + 2ax + b \\ 2a = 3 \\ b = 4 \\ a = \frac{3}{2} \\ b = 4$

You might be interested in

Choose the correct simplification of the expression (2x2y6z5)(5x4y5z3).

erastovalidia [21]

Answer:

The correct option is 4.

Step-by-step explanation:

The given expression is

$(2x^2y^6z^5)(5x^4y^5z^3)$

Simplify the given expression.

$(2x^2y^6z^5)\times (5x^4y^5z^3)$

$(2\times 5)(x^2x^4)(y^6y^5)(z^5z^3)$

$10x^{2+4}y^{6+5}z^{5+3}$ $[\because x^mx^n=x^{m+n}]$

$10x^6y^{11}z^8$

Therefore correct option is 4.

8 0

3 years ago

Which rational number is positive

victus00 [196]

Answer: d

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Please help!! Im confused

dimaraw [331]

Answer:

The answer is that you tutored for 4 hours.

Step-by-step explanation:

There are 6 total squares, so then you must divide 12 by the number of squares (6), to find the value of one square. Then, you would find that one square equals 2 hours, and since you have 2 squares that means that you tutored for 4 hours.

Hope it helps!

5 0

3 years ago

Read 2 more answers

Which values of x make this equation true?

Evgesh-ka [11]

Answer:

${x}^{2} + x = 12 \\ {x}^{2} + x - 12 = 0 \\ {x }^{2} + 4x - 3x - 12 = 0 \\ x(x + 4) - 3(x + 4) = 0 \\ (x + 4)(x - 3) = 0 \\ either \: \: \: or \: \\ x = - 4 \: \: \: \: \: \: \: \: \: \: \: \: x = 3$

the answer is -4 and 3

3 0

2 years ago

Is x= -2 a solution to 3x-6-12?

Ann [662]

Answer:

Actually x would be 6

4 0

3 years ago

Read 2 more answers