Help please! 6x^2+2x-9=0 A 6.4, -8.4 B 1.1, -1.4 C -2.2, 2.8 D -1.1, -1.4

1 answer:

6 0

$6x^2+2x-9=0\\\\a=6;\ b=2;\ c=-9\\\\\Delta=b^2-4ac\to\Delta=2^2-4\cdot6\cdot(-9)=220\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{-2-\sqrt{220}}{2\cdot6}=\dfrac{-2-\sqrt{220}}{12}\approx-1.4\\\\x_2=\dfrac{-2+\sqrt{220}}{2\cdot6}=\dfrac{-2+\sqrt{220}}{12}\approx1.1$

Answer: B. 1.1; -1.4.

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Explanation:
using the
substitution method
y
=
x
−
5
→
(
1
)
y
=
−
2
x
+
4
→
(
2
)
since both equations are expressed in terms of x we

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x
−
5
=
−
2
x
+
4
add 2x to both sides
2
x
+
x
−
5
=
−
2
x
+
2
x
+
4
⇒
3
x
−
5
=
4
add 5 from both sides
3
x
+
5
−
5
=
4
+
5
⇒
3
x
=
9
divide both sides by 3
3
x
3
=
9
3
⇒
x
=
3
substitute this value in
(
1
)
y
=
3
−
5
=
−
2
As a check
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(
2
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3 years ago

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y = 2x + 2 ==> -2x + y = 2

y = 2x + 5 ==> -2x + y = 5

This tells us the parallelogram in the x-y plane corresponds to the rectangle in the u-v plane with 1 ≤ u ≤ 4 and 2 ≤ v ≤ 5.

Compute the Jacobian determinant for this change of coordinates:

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