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3 years ago

Whats an example of a quadratic equation that cannot be factored with a real solution?

Mathematics

1 answer:

Makovka662 [10]3 years ago

4 0

Answer:

Step-by-step explanation:

The key here is the "discriminant." That's b^2 - 4ac, where a, b and c are the coefficients of the quadratic.

If the discriminant turns out to be positive, we have two real, different roots.

If the discriminant turns out to be zero, we have two real, equal roots.

Finally, if the discriminant turns out to be negative, we have two complex, different roots.

Just supposing that a = 1 and b = 2, the discriminant would be

2^2 - 4(1)c. Set this equal to 0 and solve for c: 4 - 4c = 0, or c = 1. If c is greater than 1, the discriminant will be negative and we will have two comlex, different roots.

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Solve for 6y+12=3y-3

kherson [118]

6y + 12 = 3y - 3

Subtract 3y on both sides

3y + 12 = -3

Subtract 12 on both sides

3y = -15

Divide by 3 on both sides

y = -5

5 0

3 years ago

Read 2 more answers

PLEASE HELP URGENT THANK YOU

Fynjy0 [20]

So far the only true statement I see is that function f is increasing whilst function g is decreasing.

I cannot say yes to the first one as there is a lack of x-intercepts.

Past the interval of 0,2, there are no changes in the line of function g as it remains the same despite the increase in y. Function f on the other hand has a rate of change.

The y-intercepts are the same (positive 2) so the third statement is out.

8 0

2 years ago

Which function in vertex form is equivalent to f(x) = x² + x +1?

mylen [45]

Answer:

$\left( x+\frac{1}{2} \right)^{2} + \frac{3}{4}$

Step-by-step explanation:

NOTE :

$x^2+ax=\left( x+\frac{a}{2} \right)^{2} -\left( \frac{a}{2} \right)^{2}$

………………………………………

f(x) = x² + x +1

$=x^{2}+2\times \frac{x}{2} +1$

$=\left( x+\frac{1}{2} \right)^{2} -\left( \frac{1}{2} \right)^{2} +1$

$=\left( x+\frac{1}{2} \right)^{2} - \frac{1}{4}+1$

$=\left( x+\frac{1}{2} \right)^{2} + \frac{3}{4}$

4 0

2 years ago

What's 5.78,-5 7/8,-5.9 from least to greatest?

blagie [28]

-5.9, -5 7/8, 5.78 because negative is less than zero and 7/8= .850 :)

3 0

3 years ago

1. The cost of 5 squash and 2 zucchini is $1.32. Three squash and 1 zucchini cost $0.75. Find the cost of each vegetable.

kogti [31]

Squash = s
zucchini = z

5s+2z = 1.32 ---(1)
3s+z = 0.75 ----(2)

2*(2); 6s+2z = 1.50 ---(3)
(3)-(1); s = 1.50-1.32= 0.18
then 3s + z = 0.75
3(0.18) + z = 0.75
0.54 + z = 0.75
so z = 0.75 - 0.54 = 0.21

so that the cost of squash is $0.18 and the cpst of zucchini is $0.21

4 0

3 years ago