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2 years ago

9

What is the value discriminate b squared -4 AC for the quadratic equation zero equals negative 2X squared minus 3X +8 and what d

oes it mean about the number of real solutions equation has

1 answer:

SVEN [57.7K]2 years ago

4 0

Answer:

Step-by-step explanation:

-2x²-3x+8=0

disc.=b²-4ac=(-3)²-4(-2)(8)=9+64=73>0

it has two real irrational solutions.

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MAXImum [283]

Alright, so the first thing I would do is find the LCM of the dividends, which is 20. So we have 10 15/20 and 6 16/20. You can either leave it there and just carry the one (20 in this case) for 3 19/20.
If you multiply 10 by 20 and add 15, or 6 by 20 and add 16 (which I think is easier) you get 215/20 - 136/20 = 79/20. It can be simplified from here if you teacher wants it in the future.

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For what value(s) of k will the relation not be a function?<br> A = {(3k−4, 16), (4k, 32)}

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We won't have a function if for same value of x in (x,y) we get different values y.

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3k-4=4k | solve for k

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no check the values y for the elements of the relation

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Let me know if you have any questions.

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Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

<u>case A)</u> $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

<u>case B)</u> $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

<u>case C)</u> $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

<u>case D)</u> $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

<u>the answer is</u>

$f(x)=(x-2)(x+2)$

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Answer:

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