Answer:
a = 1, -3
Step-by-step explanation:
Given equation:
[tex](a-6)(a+8)=-45[/tex]
Expand the brackets:
[tex]\implies a^2+8a-6a-48=-45[/tex]
[tex]\implies a^2+2a-48=-45[/tex]
Add 45 to both sides:
[tex]\implies a^2+2a-48+45=-45+45[/tex]
[tex]\implies a^2+2a-3=0[/tex]
To factor a quadratic in the form [tex]ax^2+bx+c[/tex], find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex].
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:
[tex]\implies a^2+3a-a-3=0[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies a(a+3)-1(a+3)=0[/tex]
Factor out the common term (a + 3):
[tex]\implies (a-1)(a+3)=0[/tex]
Apply the zero product property:
[tex]\implies (a-1)=0 \implies a=1[/tex]
[tex]\implies (a+3)=0 \implies a=-3[/tex]
Verify the solutions by inputting the found values of a into the original equation:
[tex]a=1 \implies (1-6)(1+8) & =-5 \cdot 9 = -45[/tex]
[tex]a=-3 \implies (-3-6)(-3+8) & =-9 \cdot 5 = -45[/tex]
Hence both found values of a are solutions of the given equation.
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