Question

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Answer 1

Answer:

a = 1,  -3

Step-by-step explanation:

Given equation:

$(a-6)(a+8)=-45$

Expand the brackets:

$\implies a^2+8a-6a-48=-45$

$\implies a^2+2a-48=-45$

Add 45 to both sides:

$\implies a^2+2a-48+45=-45+45$

$\implies a^2+2a-3=0$

To factor a quadratic in the form $ax^2+bx+c$, find two numbers that multiply to $ac$ and sum to $b$.  

Two numbers that multiply to -3 and sum to 2 are: 3 and -1.  

Rewrite the middle term as the sum of these two numbers:

$\implies a^2+3a-a-3=0$

Factorize the first two terms and the last two terms separately:

$\implies a(a+3)-1(a+3)=0$

Factor out the common term (a + 3):

$\implies (a-1)(a+3)=0$

Apply the zero product property:

$\implies (a-1)=0 \implies a=1$

$\implies (a+3)=0 \implies a=-3$

Verify the solutions by inputting the found values of a into the original equation:

$a=1 \implies (1-6)(1+8) & =-5 \cdot 9 = -45$

$a=-3 \implies (-3-6)(-3+8) & =-9 \cdot 5 = -45$

Hence both found values of a are solutions of the given equation.

Learn more about factorizing quadratics here:

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Answer 2

Answer:

$a = 1, \ - 3$

Explanation:

$(a-6)(a+8) = -45$

distribute

$a^2 + 8a - 6a - 48 = -45$

collect terms

$a^2 + 8a - 6a - 48 + 45=0$

simplify

$a^2 + 2a - 3=0$

factor

$(a - 1)(a + 3)= 0$

set to zero

$a = 1, \ - 3$