Solve for x in the equation x squared minus 10 x + 25 = 35. x = 5 plus-or-minus 2 StartRoot 5 EndRoot x = 5 plus-or-minus StartRoot 35 EndRoot x = 10 plus-or-minus 2 StartRoot 5 EndRoot x = 10 plus-or-minus StartRoot 35 EndRoot

Given the quadratic equation:

x^2 - 10x + 25 = 35

Solve for x.

Solution:

Equate all terms on one side, hence we transpose 35 on the other side of the equation. We must take note that in transposing a number, the sign changes.

x^2 - 10x + 25 - 35 = 0

x^2 - 10x - 10 = 0

a = 1

b = -10

c = -10

In order to solve for x we simply must look at the 2nd and 3rd value. First we must think of two numbers that when added, the answer is -10, and when multiplied, the answer is -10. Hence, if we are unable to find the number, we will use the quadratic formula.

The Quadratic formula:

x = −b ± √(b^2 − 4ac)/2a

is used to solve quadratic equations where a ≠ 0, in the form
ax^2+bx+c=0

When b^2−4ac=0 there is one real root.

When b^2−4ac>0 there are two real roots.

When b^2−4ac<0 there are no real roots, only a complex number.

Substitute the given values of a, b and c to the quadratic formula.

x = −b ± √(b^2 − 4ac)/2a
x = −(-10) ± √((-10)^2 − 4(1)(-10))/2(1)
x = 10 ± √(100 + 40)/2
x = 10 ± √(140)/2
x = 10 ± 11.83/2

Solve for + - separately.

x = 10 + 11.83 /2
x = 21.83/2
x =10.95

x = 10 - 11.83/2
x = -1.83/2
x = -0.915

Final answer:

x = 10.95
x = -0.915