# How do you find the zeros of #f(x)=5x^2-25x+30#?

See a solution process below:

We can factor this function as:

Solution 1:

Solution 2:

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To find the zeros of ( f(x) = 5x^2 - 25x + 30 ), you need to set the function equal to zero and solve for ( x ) using the quadratic formula or factoring:

[ 5x^2 - 25x + 30 = 0 ]

You can either factor the quadratic expression or use the quadratic formula:

Factoring: [ 5x^2 - 25x + 30 = 5(x^2 - 5x + 6) = 5(x - 2)(x - 3) ]

So the zeros are ( x = 2 ) and ( x = 3 ).

Quadratic Formula: [ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ]

For ( f(x) = 5x^2 - 25x + 30 ), where ( a = 5 ), ( b = -25 ), and ( c = 30 ):

[ x = \frac{{-(-25) \pm \sqrt{{(-25)^2 - 4(5)(30)}}}}{{2(5)}} ] [ x = \frac{{25 \pm \sqrt{{625 - 600}}}}{{10}} ] [ x = \frac{{25 \pm \sqrt{{25}}}}{{10}} ] [ x = \frac{{25 \pm 5}}{{10}} ]

So the solutions are ( x = \frac{{25 + 5}}{{10}} = \frac{{30}}{{10}} = 3 ) and ( x = \frac{{25 - 5}}{{10}} = \frac{{20}}{{10}} = 2 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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