Write a quadratic equation in standard form with the given roots

Mathematics

1 answer:

stich3 [128]2 years ago

5 0

${x}^{2} - \frac{15x}{2} + \frac{7}{2}$

Step-by-step explanation:

So we go backwards, since we have x= 1/2 , 7

$(x - \frac{1}{2} )(x - 7)$

Now we have to foil to get the quadratic

${x}^{2} - 7x - \frac{1}{2} x + \frac{7}{2}$

We must simplify the terms

${x}^{2} - \frac{14x}{2} - \frac{x}{2} + \frac{7}{2}$

Our final result comes out to be

${x}^{2} - \frac{15x}{2} + \frac{7}{2}$

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In the third degree polynomial function f(x) =x^3 + 4x, why are 2i and -2i zeros?

IgorC [24]

Answer:

We just need to evaluate and get f(2i)=0, f(-2i)=0.

Step-by-step explanation:

Since $i^2=-1$ , then $i^3=i^2i=-i$ , and we can apply this when we evaluate $f(x) =x^3 + 4x$ for 2i and -2i.

First we have:

$f(2i) =(2i)^3 + 4(2i)=2^3i^3+8i=8(-i)+8i=0$

Which shows that 2i is a zero of f(x).

Then we have:

$f(-2i) =(-2i)^3 + 4(-2i)=(-2)^3i^3-8i=-8(-i)-8i=8i-8i=0$

Which shows that -2i is a zero of f(x).

4 0

3 years ago

How do I solve this?

Ivenika [448]

$\overline{XS}\cong\overline{YT}\Rightarrow 3m+7=4.2m+5\ \ \ |-7\\\\3m=4.2m-2\ \ \ \ |-4.2m\\\\-1.2m=-2\ \ \ \ |:(-1.2)\\\\m=\dfrac{2}{1.2}\\\\m=\dfrac{20}{12}\\\\m=\dfrac{5}{3}\to m=1\dfrac{2}{3}$

$\overline{YS}\cong\overline{XT}\Rightarrow3\dfrac{1}{2}r+2=2r+5\ \ \ \ |-2\\\\3\dfrac{1}{2}r=2r+3\ \ \ \ |-2r\\\\1\dfrac{1}{2}r=3\ \ \ |\cdot2\\\\3r=6\ \ \ \ |:3\\\\r=2$