How do I solve this equation and check the solutions?

Mathematics

1 answer:

bearhunter [10]3 years ago

5 0

Answer:

{-6}

Step-by-step explanation:

Step 1 Raise both sides of the equation to the index of the radical in the case 2

$(\sqrt{2x+13} )^{2} =(x+7)^{2}$

$2x + 13 = (x+7)(x+7)$

$2x +13 = x^{2} + 14x + 49\\0=x^{2} +12x+36$

Step 2 Factor

(x + 6)(x + 6) = 0

x + 6 = 0

x = - 6

Step 3 Check -6

$\sqrt{2(-6)+13} = -6 + 7\\\sqrt{-12+13} = 1\\\sqrt{1} = 1\\1 = 1$

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Combine Like Terms and Eliminate Parenthesis.<br>5z-[2z-(13z+8)]

katen-ka-za [31]

We need to start from the innermost parenthesis and work our way out.

The first parenthesis is $13z+8$ . These are not like terms because one involves a variable, while the other is a constant term. Two terms are summable if they involve the same power(s) of the same variable(s).

So, we can take one step outwards, and we arrive to the square brackets. We have

$2z-13z-8$

and 2z and -13z are like terms, so we can sum them:

$2z-(13z +8) = 2z-13z-8 = (2-13)z-8 = -11z-8$

Finally we arrive to the whole expression, which is

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6 0

3 years ago

the segments ab and cd are graphed on a coordinate plane. The endpoints of ab are at (9,4) and (9,20). CD is parallel to the x-a

Step2247 [10]

Answer:

21, 8

Step-by-step explanation:

The second endpoint is of the form (X, 8) since it's parallel to the x axis.

Given the congruence of the two segments, we can tell that its length is 16 units (since $|4-20| = |-16|= 16$ )

At this point the two possible endpoints for the segment are $(5-16; 8)$ or $(5+16, 8)$ . The fact that the segment has to sit in the first quadrant rules out the first option (it's endpoint will be at (-11, 8) which will set most of it in the second quadrant) and we're left with (21, 8)