# Solve the equation completing the square t^2-10t+24=0?

samhouston | Middle School Teacher | (Level 1) Associate Educator

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t^2 - 10t + 24 = 0

t^2 - 10t = -24

Take half of the coefficient of t and square it.

-10t `=>` -10 `=>` -5 `=>` 25

Add this number to both sides.

t^2 - 10t + 25 = -24 + 25

t^2 - 10t + 25 = 1

Factor the polynomial using the Perfect Square Pattern.

(t - 5)^2 = 1

Square root both sides.

t - 5  = `+-` 1

t = 6

t = 4

You can check this by graphing and finding the x-intercepts.

Notice that the x-intercepts of the parabola are 4 and 6.

See the attached email for more examples of completing the square.

ken5000 | High School Teacher | (Level 1) Adjunct Educator

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First, we move the 24 to the other side:

t^2 - 10t + 24 = 0

This is in standard form:  Ax^2 + Bx + C = 0, where A = 1, B = 10, and C = 24.

Subtract 24 from both sides.

t^2 - 10t = -24

Now we find half of B and square it.  That is what we add to both sides.  Half of -10 is -5 and (-5)^2 is 25.  So we will add 25 to both sides:

t^2 - 10t +25 = -24 + 25

Now we factor the first side.

(t-5)(t - 5) = 1

(t - 5)^2 = 1

t - 5 = 1 or t - 5 = -1

t = 6 or t = 4

The solution set is t = {-4, 6}.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To complete the square, you need to add and subtract 1 to the left side of the equation:

t^2 - 10t + 24 + 1 - 1 = 0

t^2 - 10t + 25 - 1 = 0

The first three terms are the terms of a perfect square:

(t - 5)^2 - 1 = 0

The difference of squares returns the product:

x^2 - y^2 = (x-y)(x+y)

Let x = t - 5 and y = 1

(t - 5)^2 - 1 = (t - 5 - 1)(t - 5 + 1)

(t - 5)^2 - 1 = (t-6)(t-4)

But (t - 5)^2 - 1 = 0 => (t-6)(t-4) = 0

We'll set each factor as zero:

t - 6 = 0 => t = 6

t - 4 = 0 => t = 4

The solutions of the equation are: {4 ; 6}.