v 5x - 4 x square + 3 a t t1) 4 x square - 3 x -1 (2) 3 x square + 4 x -4 (3) 5x square + 12 x + 7 (4) t cube -2t squared -15t Choose the correct answer the syllable of has been promoted to a 'beat' position between two other unstressed syllables; this does not mean, though, that it should be heavily stressed in reading aloud. Where promotion occurs on the last syllable of an iambic line, it sometimes produces a weak ending.the syllable of has been promoted to a 'beat' position between two other unstressed syllables; this does not mean, though, that it should be heavily stressed in reading aloud. Where promotion occurs on the last syllable of an iambic line, it sometimes produces a weak ending.v 5x - 4 x square + 3 a t t1) 4 x square - 3 x -1 (2) 3 x square + 4 x -4 (3) 5x square + 12 x + 7 (4) t cube -2t squared -15t Choose the correct answer the syllable of has been promoted to a 'beat' position between two other unstressed syllables; this does not mean, though, that it should be heavily stressed in reading aloud. Where promotion occurs on the last syllable of an iambic line, it sometimes produces a weak ending.the syllable of has been promoted to a 'beat' position between two other unstressed syllables; this does not mean, though, that it should be heavily stressed in reading aloud. Where promotion occurs on the last syllable of an iambic line, it sometimes produces a weak ending.
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Answers & Comments
Answer:
Explanation:
Step 1. Write your equation in standard form.
5
x
2
+
2
x
−
3
=
0
Step 2. Move the constant to the right hand side of the equation.
Add
3
to each side .
5
x
2
+
2
x
−
3
+
3
=
0
+
3
5
x
2
+
2
x
=
3
Step 3. Divide both sides of the equation by the coefficient of
x
2
.
Divide both sides by 5.
x
2
+
2
5
x
=
3
5
Step 4. Square the coefficient of x and divide by 4.
(
2
5
)
2
4
=
4
25
4
=
1
25
Step 5. Add the result to each side.
x
2
+
2
5
x
+
1
25
=
3
5
+
1
25
x
2
+
2
5
x
+
1
25
=
15
25
+
1
25
x
2
+
2
5
x
+
1
25
=
16
25
Step 6. Take the square root of each side.
x
+
1
5
=
±
4
5
Case 1
x
1
+
1
5
=
+
4
5
x
1
=
4
5
−
1
5
=
4
−
1
5
x
1
=
3
5
Case 2
x
2
+
1
5
=
−
4
5
x
2
=
−
4
5
−
1
5
=
−
4
−
1
5
=
−
5
5
x
2
=
−
1
So
x
=
3
5
or
x
=
−
1
Check: Substitute the values of
x
back into the quadratic.
(a)
x
=
3
5
5
x
2
+
2
x
−
3
=
5
(
3
5
)
2
+
2
(
3
5
)
−
3
=
5
(
9
25
)
+
6
5
−
3
=
9
5
+
6
5
−
15
5
=
9
+
6
−
15
5
=
0
.
(b)
x
=
−
1
5
x
2
+
2
x
−
3
=
5
(
−
1
)
2
+
2
(
−
1
)
−
3
=
5
(
1
)
–
2
−
3
=
5
−
2
−
3
=
0
Answer:
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