b^{2}+5=8b−10

pedzenekO

pedzenekO

Answered question

2020-11-26

b2+5=8b10

Answer & Explanation

Gennenzip

Gennenzip

Skilled2020-11-27Added 96 answers

To solve a quadratic equation, we need to first move all the terms to one side.
Subtracting 8b on both sides of b2+5=8+5=8b10 gives b28b+5=10.
Adding 10 on both sides then gives b28b+15=0.
Now that we have all the terms on one side, we want to see if the quadratic is factorable. If it's not factorable, we will need to use the Quadratic Formula to find the solutions.
A quadratic of the form x2+Bx+Cx can be factored to (x+m)(x+n) if there are two numbers, m and n, such that m+n=B and mn=C.
For b28b+15, we then need to see if there are two numbers that multiply to 15 and add to −8. Since 5(3)=15 and 5+(3)=8, then the two numbers will be −5 and −3.
b28b+15b can then factor to (b−5)(b−3).
The equation b28b+15=0 can then be written as (b−5)(b−3)=0.
The Zero Product Property states that if ab=0, then a=0 or b=0. Using this property, if (b5)(b3)=0, we can then say that b−5=0 or b−3=0.
Solving b−5=0 for bb gives b=5. Solving b−3=0 for b gives b=3.
The two solutions of the equation are then b=3 and b=5.

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