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Maksim231197 [3]
3 years ago
15

Which of the following is not a condition that must be met before you can use the quadratic formula to find the solutions of an

equation?
A. The coefficient of the x2-term can't be zero.
B. There can be no term whose degree is higher than 2.
C. One side of the equation must be zero.
D. The coefficient of the x2-term must be positive.
Mathematics
1 answer:
ivann1987 [24]3 years ago
6 0

Answer:

"The coefficient of x²-term must be positive"

D is correct

Step-by-step explanation:

Given: y=ax^2+bx+c

We are given a quadratic equation. The quadratic equation whose degree is 2.

The leading coefficient of quadratic can't be 0.

y=ax^2+bx+c

a\neq 0

The highest degree should be 2

For quadratic formula:

ax^2+bx+c=0

First we make right side 0 and then apply quadratic formula.

Hence, The condition which is not true for quadratic formula "The coefficient of x²-term must be positive"

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erma4kov [3.2K]

Answer:

Correct options are 2, 5 and 7.

Step-by-step explanation:

Consider the given vertices of triangle are A(-3,-3), B(-3,2) and C(1,2).

Distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Using distance formula, we get

AB=\sqrt{(-3-(-3))^2+(2-(-3))^2}

AB=\sqrt{(0)^2+(5)^2}

AB=\sqrt{25}

AB=5

Similarly,

BC=\sqrt{(1-(-3))^2+(2-2)^2}=4

AC=\sqrt{(1-(-3))^2+(2-(-3))^2}=\sqrt{16+25}=\sqrt{41}

From the above calculation it is clear that AC>AB and AC>BC.

According to Pythagoras theorem, in a right angle triangle, the square of largest side is equal to the sum of squares of two small sides.

hypotenuse^2=base^2+perpendicular^2

AC^2=(\sqrt{41})^2=41

AB^2+BC^2=(5)^2+4^2=24+16=41=AC^2

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Therefore, the correct options are 2, 5 and 7.

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Answer:

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3 years ago
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