If a number is added to its square, the result is 56. How do you find the number?

2 Answers
Mar 29, 2018

Answer:

#color(blue)(a=-8)#

#color(blue)(a=7)#

Explanation:

Let the number be #a#:

#:.#

#a^2+a=56#

We now solve the equation:

#a^2+a-56=0#

Factor:

#(a+8)(a-7)=0=>a=-8 and a=7#

Mar 29, 2018

Answer:

Call the number #n#, then we know #n+n^2=56#. We can write this as a quadratic equation in standard form: #n^2+n-56=0#. Using the quadratic formula, or by factorising, we find #n=-8# or #7#..

Explanation:

I normally use the quadratic formula: #n=(-b+-sqrt(b^2-4ac))/(2a)#

It's not hard to remember, and it always works.

For this particular one, though, I will factorise instead. I need two numbers that multiply to give #-56# and add to give #+1#. Two numbers that fit are #8# and #-7#.

So #(n+8)(N-7)=0#

If you expand that you will get back our quadratic.

The two 'roots' - the solutions of the quadratic - are the values of #n# that make each bracket go to zero.

Those roots are #-8# and #7#. Either value fulfils the criteria in the question.