Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
Mid-Chapter Quiz
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Exercise 21 Page 575

Make sure you rewrite the equation leaving all the terms on one side, and that you factor out the greatest common factor if it exists.

7 and - 4

Practice makes perfect

We want to solve the given equation by factoring.

Factoring

Let's start by writing all the terms on one side of the equals sign.
t^2-3t=28
t^2-3t-28=0
Next, let's rewrite the middle term (- 3x) as two terms ( b_1x+ b_2x). Since the expression matches the form ax^2+bx+c, we know the following. b_1 * b_2 = ac = - 28 and b_1 + b_2 = b = - 3 Because ac is negative we know the two terms must have opposite signs, and because b is negative we know the negative term must be larger than the positive. c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result 2 &- 14 &2 + (-14) &- 12 4 & - 7 & 4 + ( - 7) &- 3 Now, we can rewrite the equation by substituting in the values we determined above. t^2-3t-28 = 0 ⇕ t^2 + 4t - 7t -28 = 0 Finally, we can finish factoring the equation.
t^2+4t-7t-28=0
â–Ľ
Factor out t & - 7
t(t+4)-7t-28=0
t(t+4)-7(t+4)=0
(t+4)(t-7)=0

Solving

To solve this equation, we will apply the Zero Product Property.
(t+4)(t-7)=0
lct+4=0 & (I) t-7=0 & (II)
lt=- 4 t-7=0
lt=- 4 t=7
We found that the solutions to the equation are 7 and - 4.

Checking Our Answer

Checking our answer
We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with t=- 4.
t^2-3t=28
( -4)^2-3( -4)? =28
â–Ľ
Simplify
16-3(-4)? =28
16-(- 12)? =28
28=28 âś“
Substituting and simplifying created a true statement, so we know that t=- 4 is a solution of the equation. Let's move on to t=7.
t^2-3t=28
( 7)^2-3( 7)? =28
â–Ľ
Simplify
49-3(7)? =28
49-21? =28
28=28 âś“
Again, we created a true statement. t=7 is indeed a solution of the equation.