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polet [3.4K]
3 years ago
9

Can you show us how to find the discriminant of the quadratic x^2 + 2x -2 =0

Mathematics
2 answers:
vlabodo [156]3 years ago
6 0
the\ discriminant\ of\ the\ quadratic\ ax^2+bx+c=0\\\\\Delta=b^2-4\cdot a\cdot c\\-------------------------\\\\ x^2 + 2x -2 =0\\\\\Delta=2^2-4\cdot1\cdot(-2)=4+8=12\\\\discriminant=12
lorasvet [3.4K]3 years ago
4 0
<em>Step #1: </em>
Make sure the equation is in the form of [ Ax² + Bx + C = 0 ].

Yours is already in that form.
A = 1
B = 2
C = -2

<em>Step #2:</em>
The 'discriminant' for that equation is [ B² - 4 A C ].
That's all there is to it, but it can tell you a lot about the roots of the equation.

-- If the discriminant is zero, then the left  side of the equation is a perfect square,
and both roots are equal. 

-- If the discriminant is greater than zero, the the roots are real and not equal.

-- If the discriminant is less than zero, then the roots are complex numbers.

The discriminant of your equation is  [ B² - 4 A C ] = 2² - 4(1)(-2) = 4 + 8 = 12

Your equation has two real, unequal roots.



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Answer:

f is increasing on interval (-infty, 13/4)

Inequality notation: x<13/4

In words: f is increasing on the interval of x that is less than 13/4.

Step-by-step explanation:

f is increasing on interval of x if f' of such interval is positive.

f=-2x^2+13x-8

Differentiate both sides

(f)'=(-2x^2+13x-8)'

Sum and difference rule:

f'=(-2x^2)'+(13x)'-(8)'

Constant multiple rule:

f'=-2(x^2)'+13(x)'-(8)'

Power rule (recall x=x^1):

f'=-2(2x^1)+13(1x^0)-(8)'

Constant rule:

f'=-2(2x^1)+13(1x^0)-(0)

Recall again x^1=x:

f'=-2(2x)+13(1x^0)-(0)

Recall x^0=1:

f'=-2(2x)+13(1×1)-(0)

Associative property of multiplication:

f'=-(2×2)x+13(1×1)-(0)

Performed grouped multiplication:

f'=-(4)x+13(1)-(0)

f'=-4x+13-(0)

Additive identity:

f'=-4x+13

f' is positive when -4x+13>0.

Subtract 13 on both sides:

-4x>-13

Divide both sides by -4:

x<-13/-4

x<13/4

f is increasing on interval (-infty, 13/4)

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Answer:

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Step-by-step explanation:

Since the small rectangle is ½ the scale drawing of the original figure, therefore the dimensions of the smaller rectangle would be half of that of the original.

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Answer:

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Step-by-step explanation:

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