Edwin and Aldrin attempt to solve a problem that reduces to a quadrati

Osvaldo Apodaca

Osvaldo Apodaca

Answered question

2021-12-11

Edwin and Aldrin attempt to solve a problem that reduces to a quadratic equation. Edwin made a mistake only in the constant term of the quadratic equation and gives answers of - 3 and - 4 for the roots. Aldrin solving the same problem made a mistake in the coefficient of the first degree term and gives answers of 2 and 5 for the roots. If you check their solutions, what would be the correct quadratic equation?

Answer & Explanation

Maricela Alarcon

Maricela Alarcon

Beginner2021-12-12Added 28 answers

Step 1
Given:
Let the quadratic equation be ax2+bx+c=0
Let the equation in which Edwin made mistake be ax2+bx+c1=0
And roots of this equation are: -3 and -4
We know that the product of roots =(constant term)(coefficient of x2}
So,
For Edwin’s equation:
3×4=c1a
1) 12a=c1
The correct value of constant term (c) is 12a.
Step 2
Let the equation in which Aldrin made mistake be ax2+b1x+c=0
Roots of this equation are 2 and 5
We know that the sum of roots =(coefficient of x)(coefficient ofx2
2+5=b1a
7a=b1
The correct value of coefficient of x is (7a)
Step 3
Now put the value of coefficient of x and constant term in the quadratic equation:
ax2+(7a)x+12a=0
ax27ax+12a=0
a(x27x+12)=0
x27x+12=0
So, the correct quadratic equation is:
x27x+12=0

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