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kenny6666 [7]
3 years ago
7

What are the exact solutions of x^2 = 5x + 2?

Mathematics
1 answer:
aleksandrvk [35]3 years ago
5 0
Move everybody to one side
minus 5x+2 both sides
x^2-5x-2=0
use quadratic formula where
if you have
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}

1x^2-5x-2=0
a=1
b=-5
c=-2

x=\frac{-(-5)+/- \sqrt{(-5)^{2}-4(1)(-2)} }{2(1)}
x=\frac{5+/- \sqrt{25+8} }{2}
x=\frac{5+/- \sqrt{33} }{2}

aprox
x=5.372281323 or -0.3722813233
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Use the quadratic formula to solve the following equation -3x^2-x-3=0
tigry1 [53]

<u>Answer:</u>

x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i are two roots of equation -3 x^{2}-x-3=0

<u>Solution:</u>

Need to solve given equation using quadratic formula.

-3 x^{2}-x-3=0

General form of quadratic equation is a x^{2}+b x+c=0

And quadratic formula for getting roots of quadratic equation is

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}

Since b^{2}-4 a c is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}

8 0
3 years ago
I will give a brainlist!!!!! And a thanks help me please
kaheart [24]

Answer:


Step-by-step explanation:

question 6

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line 1

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line 2

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line 3

8 to 92

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hope that helped

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Step-by-step explanation: cause i did this

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Step-by-step explanation:

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The word sentence "A number y is no more than -8" as an inequality is:

y < -8

Therefore the variable (or unknown) y is less than -8.

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