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VLD [36.1K]
3 years ago
11

Use the quadratic formula to find both solutions to the quadratic equation given below 2x^2-3x+1=0

Mathematics
2 answers:
VLD [36.1K]3 years ago
7 0

Answer:

x_1=1\\x_2=\frac{1}{2} =0.5

Step-by-step explanation:

Given a equation of the form:

ax^2+bx+c=0

The roots of this equation can be found using the quadratic formula which is given by:

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

In this case we have this equation:

2x^2-3x+1=0

So:

a=2\\b=-3\\c=1

Using the the quadratic equation :

x= \frac{-(-3)\pm\sqrt{(-3)^{2}-4(2)(1) } }{2(2)} = \frac{3\pm\sqrt{9-8 } }{4}=\frac{3\pm 1}{4}

Therefore the two roots would be:

x_1=\frac{3+ 1}{4}=\frac{4}{4}= 1\\x_2=\frac{3- 1}{4}=\frac{2}{4}=\frac{1}{2}=0.5

kicyunya [14]3 years ago
5 0
-2   +   -1   =   -3
<span>(2x2 - 3x) + 1 = 0
</span><span>(x - 1) • (2x - 1) = 0
</span>x-1=0
x = 1<span> </span>
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