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What are the zeros of
2x² = 3x+1?

User Gel
by
6.7k points

2 Answers

5 votes

Answer:


x = (√(17) + 3)/(4) , x = (√(17) - 3)/(4)

Explanation:

Zeros of a quadratic function are the x-intercepts or solving it.

First, subtract 3x on both sides.


2x^2-3x=3x-3x+1


2x^2-3x=1

Divide both sides by 2, the value of a.


(2x^2)/(2) - (3x)/(2) = (1)/(2)


x^2 - (3)/(2) x= (1)/(2)

Add this to both sides, which makes it a perfect square:


((a)/(2) )^2

a is the coefficient of x.


((3/2)/(2) )^2


((3)/(4) )^2


(9)/(16)


x^2 - (3)/(2) x + (9)/(16) = (1)/(2) + (9)/(16)


x^2-(3)/(2) x+(9)/(16) = (17)/(16)

Factor out the left side.


(x-(3)/(4) )^2 =(17)/(16) }

Find the square root of both sides.


\sqrt{(x-(3)/(4) )^2 }} = \sqrt{(17)/(16)}


x-(3)/(4) = (√(17) )/(4) , -\frac{√(17) }4}

Add 3/4 on both sides.


x = (√(17) + 3)/(4) , x = (√(17) - 3)/(4)

User Ines Tlili
by
6.7k points
4 votes

Answer:

The zeroes of given equation is 1 and 1/2

Explanation:

2x² = 3x + 1

2x² - 3x - 1 = 3x + 1 - 3x - 1

2x² - 3x - 1 = 0

Method used:

  • Splitting the middle term

2x² - 2x - x - 1 = 0

2x(x - 1) - 1(x - 1) = 0

(x - 1) (2x - 1) = 0

x - 1 = 0 or 2x - 1 = 0

x = 1 or x = 1/2

Thus, The zeroes of given equation is 1 and 1/2

User Benjarobin
by
7.4k points
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