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How do you solve #x^2 = 2x + 15#?

Answer 1

Ethan Adkins

#x=-3# or #x=5#

#x^2=2x+15#

#hArrx^2-2x-15=0#

Splitting the middle term in #-5x# and #3x# we get

#x^2-5x+3x-15=0# or

#x(x-5)+3(x-5)=0# or

#(x+3)(x-5)=0#

Hence #x=-3# or #x=5#

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Answer 2

Genesis Allen

To solve the equation ( x^2 = 2x + 15 ), you can first rearrange it into the form ( x^2 - 2x - 15 = 0 ). Then, you can factor the quadratic expression or use the quadratic formula. Factoring may involve finding two numbers that multiply to give ( -15 ) and add to give ( -2 ). These numbers are ( -5 ) and ( 3 ). So, factoring the expression yields ( (x - 5)(x + 3) = 0 ). Setting each factor equal to zero gives ( x - 5 = 0 ) or ( x + 3 = 0 ). Solving these equations gives ( x = 5 ) or ( x = -3 ). Therefore, the solutions to the equation are ( x = 5 ) and ( x = -3 ).

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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.