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2 years ago

Find the zeros of g(x)=2x^2+32

Mathematics

1 answer:

expeople1 [14]2 years ago

4 0

Answer:

x = 4i, − 4i

Step-by-step explanation:

The roots (zeros) are the x values where the graph intersects the x-axis. To find the roots (zeros), replace y with 0 and solve for x

x = 4i, − 4i

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3 years ago

Solve for x use the completing the square method x^2+6x=5

Mrac [35]

Answer:

$x_{1} =-3 +\sqrt{14} \\\\x_{2} =-3 -\sqrt{14}$

Step-by-step explanation:

$x^{2}$ +6x-5=0

we divide the coefficient of the X by half :

in this case: 6/2 = 3 , then we do the following

The result obtained is raised to square power: 3^2=9

we sum and subtract by 9 to maintain the balance of the equation:

$x^{2}$ +6x+9-9-5=0

we have:

$(x+3)^{2}$ -9-5=0

$(x+3)^{2}$ = 14

lets apply square root on both sides of the equation:

$\sqrt{(x+3)^{2}}=\sqrt{14}$

we know:

$\sqrt{a^{2}} = abs(a)$

so we have:

abs(x+3)= $\sqrt{14}$

from where two solutions are obtained

$x_{1} + 3 =\sqrt{14} \\\\x_{2} + 3 =-\sqrt{14}$

finally we have:

$x_{1} =-3 +\sqrt{14} \\\\x_{2} =-3 -\sqrt{14}$

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3 years ago

Read 2 more answers