All categories

4 years ago

Find all the zeros of the equation

Mathematics

1 answer:

lapo4ka [179]4 years ago

6 0

Divide both sides by -3, and replace $x^2$ with $y$ . Then

$-3x^4+27x^2+1200=0\iff y^2-9y-400=0$

Factorize the quadratic in $y$ to get

$y^2-9y-400=(y+16)(y-25)=0\implies y=-16,y=25$

which in turn means

$x^2=-16,x^2=25$

But $x^2\ge0$ for all real $x$ , so we can ignore the first solution. This leaves us with

$x^2=25\implies x=\pm\sqrt{25}=\pm5$

If we allow for any complex solution, then we can continue with the solution we ignored:

$x^2=-16\implies x=\pm\sqrt{-16}=\pm i\sqrt{16}=\pm4i$

You might be interested in

Which could be the base shape of the cylinder

olga_2 [115]

Answer: i think it is a circle

Step-by-step explanation: because on the top and bottom its circle

5 0

3 years ago

Read 2 more answers

4x² + 24x + 20 common factor =

adelina 88 [10]

The common factor is 4

6 0

2 years ago

Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!

blagie [28]

Question 4

1) $\overline{BD}$ bisects $\angle ABC$ , $\overline{EF} \perp \overline{AB}$ , and $\overline{EG} \perp \overline{BC}$ (given)

2) $\angle FBE \cong \angle GBE$ (an angle bisector splits an angle into two congruent parts)

3) $\angle BFE$ and $\angle BGE$ are right angles (perpendicular lines form right angles)

4) $\triangle BFE$ and $\triangle BGE$ are right triangles (a triangle with a right angle is a right triangle)

5) $\overline{BE} \cong \overline{BE}$ (reflexive property)

6) $\triangle BFE \cong \triangle BGE$ (HA)

Question 5

1) $\angle AXO$ and $\angle BYO$ are right angles, $\angle A \cong \angle B$ , $O$ is the midpoint of $\overline{AB}$ (given)

2) $\triangle AXO$ and $\triangle BYO$ are right triangles (a triangle with a right angle is a right triangle)

3) $\overline{AO} \cong \overline{OB}$ (a midpoint splits a segment into two congruent parts)

4) $\triangle AXO \cong \triangle BYO$ (HA)

5) $\overline{OX} \cong \overline{OY}$ (CPCTC)

Question 6

1) $\angle B$ and $\angle D$ are right angles, $\overline{AC}$ bisects $\angle BAD$ (given)

2) $\overline{AC} \cong \overline{AC}$ (reflexive property)

3) $\angle BAC \cong \angle CAD$ (an angle bisector splits an angle into two congruent parts)

4) $\triangle BAC$ and $\triangle CAD$ are right triangles (a triangle with a right angle is a right triangle)

5) $\triangle BAC \cong \triangle DCA$ (HA)

6) $\angle BCA \cong \angle DCA$ (CPCTC)

7) $\overline{CA}$ bisects $\angle ACD$ (if a segment splits an angle into two congruent parts, it is an angle bisector)

Question 7

1) $\angle B$ and $\angle C$ are right angles, $\angle 4 \cong \angle 1$ (given)

2) $\triangle BAD$ and $\triangle CAD$ are right triangles (definition of a right triangle)

3) $\angle 1 \cong \angle 3$ (vertical angles are congruent)

4) $\angle 4 \cong \angle 3$ (transitive property of congruence)

5) $\overline{AD} \cong \overline{AD}$ (reflexive property)

6) $\therefore \triangle BAD \cong \triangle CAD$ (HA theorem)

7) $\angle BDA \cong \angle CDA$ (CPCTC)

8) $\therefore \vec{DA}$ bisects $\angle BDC$ (definition of bisector of an angle)

8 0

2 years ago

Can someone answer this question please please help me I really need it if it’s correct I will mark you brainliest .

Allisa [31]

Answer:

We divided the figure into 1 triangle and 2 rectangles.

Adding up the area of them, we get the total area:

A = 14 x (10 + 8 + 12 - 16)/2 + 8 x 5 + 6 x (12 + 8)

= 258 (yd2)

Hope this helps!

5 0

3 years ago

An executive in an engineering firm earns a monthly salary plus a Christmas bonus of 7100 dollars. If she earns a total of 90400

photoshop1234 [79]

90,400-7,100=83,300

83,300/12=6,941.66667

Monthly Salary Rounded: 6,941.67

5 0

3 years ago