How do you solve #x^2-10x+25=35#?
or
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#x=5+-sqrt35#
We use the quadratic formula to solve it
Where,
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Notice that the left hand side of the equation is already a perfect square trinomial.
So rather than reformulate, we can proceed as follows:
Take the square root of both ends, allowing for both positive and negative square roots to get:
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To solve the equation ( x^2 - 10x + 25 = 35 ), follow these steps:
- Subtract 35 from both sides: ( x^2 - 10x + 25 - 35 = 0 ).
- Simplify: ( x^2 - 10x - 10 = 0 ).
- Rewrite the equation in the form ( ax^2 + bx + c = 0 ): ( x^2 - 10x - 10 = 0 ).
- Use the quadratic formula: ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ).
- Substitute ( a = 1 ), ( b = -10 ), and ( c = -10 ) into the quadratic formula.
- Calculate: ( x = \frac{{-(-10) \pm \sqrt{{(-10)^2 - 4(1)(-10)}}}}{{2(1)}} ).
- Simplify inside the square root: ( x = \frac{{10 \pm \sqrt{{100 + 40}}}}{2} ).
- Further simplify: ( x = \frac{{10 \pm \sqrt{{140}}}}{2} ).
- Simplify the square root: ( x = \frac{{10 \pm 2\sqrt{{35}}}}{2} ).
- Simplify further: ( x = 5 \pm \sqrt{{35}} ).
- The solutions are ( x = 5 + \sqrt{{35}} ) and ( x = 5 - \sqrt{{35}} ).
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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.
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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