What kind of solutions does #5x^2 + 8x + 7 = 0# have?

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Answer 1 · 2018-06-15T02:23:55

Quadratic has no real solutions

The key realization is that our quadratic is of the form

#ax^2+bx+c=0#, where the value #b^2-4ac# is the discriminant, which tells us

#I.b^2-4ac >0=>2# real roots

#II.b^2-4ac<0=># no real roots

#III.b^2-4ac=0=>1# real root

In our quadratic, we know #a=5, b=8# and #c=7#. So let's plug these values in. We get

#8^2-4(5*7)#

#=>64-4(35)#

#64-140=color(blue)(-76)#

Notice, our discriminant is less than zero, so we are in scenario #II#. This means our quadratic has no real solutions.

Hope this helps!

Answer 2 · 2018-06-15T02:31:20

The roots are all complex numbers.

The quadratic equation is #x=(-b+-sqrt(b^2-4ac))/(2a)#.

Here, #a=5#, #b=8#, and #c=7#.

Substituting, we get #x=(-8+-sqrt(8^2-4*5*7))/(2*5)#.

Simplify.

#x=(-8+-sqrt(64-140))/(10)#

Here, we can already tell that the answers will be complex numbers. If you want to simplify further...

#x=(-8+-sqrt(-76))/(10)#

#x=(-8+-4sqrt(-19))/(10)#

#x=(-4+-2sqrt(-19))/(5)#