How do you use the quadratic formula to find both solutions to the quadratic equation #x^2-3x=-10#?

2 Answers
Jun 28, 2015

Answer:

Solve #y = x^2 - 3x + 10 = 0#

Explanation:

#D = d^2 = b^2 - 4ac = 9 - 40 = -31 < 0# --> #d = +- isqrt31#

There are no real roots. There are 2 complex roots.

#x = 3/2 +-( isqrt31)/2#

Jun 28, 2015

Answer:

#x^2-3x=-10# has only complex roots #x = (3+-i sqrt(31))/2#

#x^2-3x=10# has roots #x=5# and #x=-2#

Explanation:

I'm not sure the sign of #10# on the right hand side is correct, so let's deal with both cases:

Case 1 #x^2-3x=-10#

Add #10# to both sides to get #x^2-3x+10 = 0#

Let #f(x) = x^2-3x+10#.

This is of the form #ax^2+bx+c# with #a=1#, #b=-3# and #c=10#.

The discriminant #Delta# is given by the formula:

#Delta = b^2 - 4ac = (-3)^2-(4xx1xx10) = 9-40 = -31#

Since this is negative #f(x) = 0# has two distinct complex roots, given by the formula:

#x = (-b +-sqrt(Delta))/(2a) = (3+-i sqrt(31))/2#

Case 2 #x^2-3x=10#

Subtract #10# from both sides to get #x^2-3x-10 = 0#

Let #f(x) = x^2-3x-10#

This is of the form #ax^2+bx+c# with #a=1#, #b=-3# and #c=-10#

The discriminant #Delta# is given by the formula:

#Delta = b^2 - 4ac = (-3)^2-(4xx1xx-10) = 9+40 = 49 = 7^2#

This is positive and a perfect square, so the roots of #f(x) = 0# are distinct rational real numbers, given by the formula:

#x = (-b+-sqrt(Delta))/(2a) = (3+-7)/2#

That is #x = -4/2 = -2# and #x = 10/2 = 5#