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Bezzdna [24]
3 years ago
14

Use the discriminant to determine the number and type of solutions the equation has.

Mathematics
2 answers:
AysviL [449]3 years ago
6 0
ax^2+bx+c=0\ \ \ \Rightarrow\ \ \ \Delta=b^2-4\cdot a\cdot c\\\\\Delta>0\ \ \ \Rightarrow\ \ \ two\ real\ solutions\\\Delta=0\ \ \ \Rightarrow\ \ \ one\ real\ solution\\\Delta0\ \ \ \ \Rightarrow\ \ \ two\ real\ solutions\ \ \ \Rightarrow\ \ \ Ans.\ C
White raven [17]3 years ago
5 0
x^2 + 8x + 12 = 0\\ \\a=1 , b= 8, c= 12 \\ \\ \Delta =b^2-4ac = 8^2 -4\cdot1\cdot12= 64-48= 16\\ \\\Delta > 0 \\ \\ Answer : \ C. \ \ two \ rational \ solutions


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Answer:

x = 2, y = 4

Step-by-step explanation:

x + \frac{3}{4} y = 5 ( multiply through by 4 to clear the fraction )

4x + 3y = 20 → (1)

x - \frac{1}{2} y = 0 ( multiply through by 2 to clear the fraction )

2x - y = 0 → (2)

Multiplying (2) by 3 and adding to (1) will eliminate y

6x - 3y = 0 → (3)

Add (1) and (3) term by term to eliminate y

10x + 0 = 20

10x = 20 ( divide both sides by 10 )

x = 2

Substitute x = 2 into either of the 2 equations and solve for y

Substituting into (1)

4(2) + 3y = 20

8 + 3y = 20 ( subtract 8 from both sides )

3y = 12 ( divide both sides by 3 )

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Solution is x = 2, y = 4

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