How do you solve #x^2-10x+25=35#?

Answer 1

#x = 5 + sqrt(35)#
or
#x = 5 - sqrt(35)#

Take 35 from both sides so that the quadratic equation is equal to 0. #x^2 -10x - 10 = 0#
Solve for x using the following equation: #x = (-b +- sqrt(b^2 -4ac))/(2a)# #= (-(-10) +- sqrt((-10)^2 -4(1)(-10)))/(2(1))# #= (10 +- sqrt(100 +40))/2# #= (10 +- sqrt(140))/2# #= (10 +- (sqrt(35)*sqrt(4)))/2# #= (2(5) +- 2(sqrt(35)))/2# #= (2(5 +- sqrt(35)))/2# #= 5 +- sqrt(35)#
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Answer 2

#x=5+-sqrt35#

#color(blue)(x^2-10x+25=35#
#rarrx^2-10x+25-35=0#
#rarrx^2-10x-10=0#
This is a quadratic equation (in form of #color(orange)(ax^2+bx+c=0#)

We use the quadratic formula to solve it

#color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)#

Where,

#color(violet)(a=1#
#color(violet)(b=-10#
#color(violet)(c=-10#
#rarrx=(-(-10)+-sqrt((-10)^2-4(1)(-10)))/(2(1))#
#rarrx=(10+-sqrt(100-(-40)))/(2)#
#rarrx=(10+-sqrt(100+40))/(2)#
#rarrx=(10+-sqrt(140))/(2)#
#rarrx=(10+-sqrt(4*35))/(2)#
#rarrx=(10+-2sqrt(35))/(2)#
#rarrx=(cancel10^5+-cancel2^1sqrt(35))/(cancel2^1)#
#rArrcolor(green)(x=5+-sqrt35#
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Answer 3

#x = 5+-sqrt(35)#

Notice that the left hand side of the equation is already a perfect square trinomial.

So rather than reformulate, we can proceed as follows:

#(x-5)^2 = x^2-10x+25 = 35#

Take the square root of both ends, allowing for both positive and negative square roots to get:

#x-5 = +-sqrt(35)#
Add #5# to both sides to find:
#x = 5+-sqrt(35)#
Finally note that #35=5*7# has no square factors, so #sqrt(35)# is already in simplest form.
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Answer 4

To solve the equation ( x^2 - 10x + 25 = 35 ), follow these steps:

  1. Subtract 35 from both sides: ( x^2 - 10x + 25 - 35 = 0 ).
  2. Simplify: ( x^2 - 10x - 10 = 0 ).
  3. Rewrite the equation in the form ( ax^2 + bx + c = 0 ): ( x^2 - 10x - 10 = 0 ).
  4. Use the quadratic formula: ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ).
  5. Substitute ( a = 1 ), ( b = -10 ), and ( c = -10 ) into the quadratic formula.
  6. Calculate: ( x = \frac{{-(-10) \pm \sqrt{{(-10)^2 - 4(1)(-10)}}}}{{2(1)}} ).
  7. Simplify inside the square root: ( x = \frac{{10 \pm \sqrt{{100 + 40}}}}{2} ).
  8. Further simplify: ( x = \frac{{10 \pm \sqrt{{140}}}}{2} ).
  9. Simplify the square root: ( x = \frac{{10 \pm 2\sqrt{{35}}}}{2} ).
  10. Simplify further: ( x = 5 \pm \sqrt{{35}} ).
  11. The solutions are ( x = 5 + \sqrt{{35}} ) and ( x = 5 - \sqrt{{35}} ).
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Answer from HIX Tutor

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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