Question

Help!!Hurry!!!! NO SPAM!!!!!!!!! I will give brainliest if you get it correct, and show your work. √(5x+21)=x+3

Answer 1

Answer:

Step-by-step explanation:

$\sqrt{5x+21}=x+ 3 \\\\Take \ square \\\\\\5x + 21 = (x + 3)^{2}$

5x + 21 = x² + 2*x*3 + 3²

5x + 21 = x² +6x + 9

x² + 6x+9 -5x - 21 = 0

x² + 6x - 5x +  9 -21 = 0

x² + x - 12 = 0

Sum = 1

Product = -12

Factors = 4 , -3    {4*(-3) = -12  and 4 + (-3) = 1}

x² + 4x - 3x - 12 = 0

x(x + 4) - 3(x +4) = 0

(x + 4)(x - 3) =0

x = - 4 , 3

Answer 2

Answer:

$\boxed{\bf x = 3}$

Step-by-step explanation:

Equation Given:

$\sf \longmapsto \sqrt{(5x+21)}=x+3$

Solution:-

Square both sides and remove square root:

$\sf \longmapsto \: 5x + 21 =( x + {3})^{2}$

$\sf \longmapsto \: 5x + 21 = {x}^{2} + 6x + 9$

Subtract x^2+6x+9 from both sides :

$\sf \longmapsto5x+21−(x {}^{2} +6x+9)=x {}^{2} +6x+9−(x {}^{2} +6x+9)$

$\sf \longmapsto5x+21−(x {}^{2} +6x+9) = 0$

$\sf \longmapsto \: - x {}^{2} - x + 12 = 0$

Factor LHS of the equation:

$\sf \longmapsto( - x + 3) \times (x + 4) = 0$

Put factors = to 0:

$\sf \longmapsto \: - x + 3 = 0 \: \rm \: or \: \: \sf \: x = - 4$

$\sf \longmapsto \: x = 3 \: \rm \: or \: \sf \: x = - 4$

$\boxed{\sf x = 3 \: \rm \: or \: \sf \: x = - 4}$

Let's check.

First x=3 :

Put the values on the given Equation According to the question to verify:

$\sf \longmapsto \: \sqrt{5 \times 3 + 21} = 3 + 3$

$\sf \longmapsto \sqrt{15 + 21} = 6$

$\sf \longmapsto \sqrt{36[6×6]} = 6$

$\sf \longmapsto6 = 6$(True)

Hence, this equation is true when x = 3.

Now x=-4:

Put the values on the given Equation According to the question to verify:

$\sf \longmapsto \sqrt{5 \times - 4 + 21} = - 4 + 3$

$\sf \longmapsto \sqrt{ - 20 + 21} = - 1$

$\sf \longmapsto \sqrt{1} = - 1$

$\sf \longmapsto1 = - 1$(False)

The equation is false when x = -4 is plugged, so the actual answer will be x = 3 where the equation comes true after putting the values.

______________________________________

I hope this helps!