How do you solve #x^2-10x+18=0# using the quadratic formula?

Answer 1

You simply put the coefficients of the equation into the formula and solve it algebraically.

For #ax^2 + bx + c = 0#, the values of x which are the solutions of the equation are given by: #x = (−b±√[b^2−4ac])/(2a)#

In this case, a = 1, b = -10 and c = 18

#x = (−(-10) ± sqrt[(-10^2 − 4*1*18]))/(2*1)#

#x = (10 ± sqrt[(100 – 72)])/2# ; #x = (10 ± sqrt[28])/2#

#x = (10 ± 5.29)/2#

x = 2.35, and x = 7.645

CHECK: 7.645^2 – 10*7.645 + 18 = 0 ; 58.446 – 76.45 + 18 = 0 ; 0 = 0 correct. See https://tutor.hix.ai for instructions.

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Answer 2

Savannah Anderson

To solve the quadratic equation (x^2 - 10x + 18 = 0) using the quadratic formula:

[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}]

Where (a = 1), (b = -10), and (c = 18):

[x = \frac{{-(-10) \pm \sqrt{{(-10)^2 - 4 \cdot 1 \cdot 18}}}}{{2 \cdot 1}}]

[x = \frac{{10 \pm \sqrt{{100 - 72}}}}{{2}}]

[x = \frac{{10 \pm \sqrt{{28}}}}{{2}}]

Therefore, the solutions are:

[x = \frac{{10 + \sqrt{{28}}}}{{2}}] or [x = \frac{{10 - \sqrt{{28}}}}{{2}}]

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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.