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If -1/2 and -2 are the two roots of the quadratic equation ax^2 + 5
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08 Jun 2022, 02:00
08 Jun 2022, 02:00
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If \(-\frac{1}{2}\) and \(-2\) are the two roots of the quadratic equation \(ax^2 + 5x + b = 0\), then what is the value of a and b?
A. a = -2 and b = -2
B. a = -2 and b = -1
C. a = -2 and b = 1
D. a = 2 and b = 1
E. a = 2 and b = 2
A. a = -2 and b = -2
B. a = -2 and b = -1
C. a = -2 and b = 1
D. a = 2 and b = 1
E. a = 2 and b = 2
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022)
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GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
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Re: If -1/2 and -2 are the two roots of the quadratic equation ax^2 + 5
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08 Jun 2022, 05:28
08 Jun 2022, 05:28
Carcass wrote:
If -1/2 and -2 are the two roots of the quadratic equation \(ax^2 + 5x + b = 0\), then what is the value of a and b?
A. a = -2 and b = -2
B. a = -2 and b = -1
C. a = -2 and b = 1
D. a = 2 and b = 1
E. a = 2 and b = 2
A. a = -2 and b = -2
B. a = -2 and b = -1
C. a = -2 and b = 1
D. a = 2 and b = 1
E. a = 2 and b = 2
Key concept: If \(x = -1/2\) and \(x = -2\) are roots of the equation, then those values SATISFY the equation.
Let's plug \(x = -0.5\) and \(x = -2 \) into the equation...
If \(x = -0.5\), we get: \(a(-0.5)^2 + 5(-0.5) + b = 0\)
Simplify: \(0.25a - 2.5 + b = 0\)
Take this prettier, let's multiply both sides by \(4\) to get: \(a - 10 + 4b = 0\)
Add \(10\) to both sides: \(a + 4b = 10\)
If \(x = -2\), we get: \(a(-2)^2 + 5(-2) + b = 0\)
Simplify: \(4a - 10 + b = 0\)
Add \(10\) to both sides: \(4a + b = 10\)
We now have the following system of equations:
\(a + 4b = 10\)
\(4a + b = 10\)
Solve to get: \(a = 2\) and \(b = 2\)
Answer: E
Re: If -1/2 and -2 are the two roots of the quadratic equation ax^2 + 5
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10 Jun 2022, 05:46
10 Jun 2022, 05:46
1
There are numerous ways to solving this if you understand how roots are obtained from a quadratic equation, but since we are given the roots, we can simply plug in and solve two equation for one unknown:
Substitute x=-1/2 and x=-2
4a - 10 + b = 0 ------------ (1)
1/4a - 5/2 + b = 0 -------- (2)
(15/4)a = 15/2 ---> a = 2
Now substituting the value of a back in equation (1) or (2), we get b = 2
The answer is (E)
Substitute x=-1/2 and x=-2
4a - 10 + b = 0 ------------ (1)
1/4a - 5/2 + b = 0 -------- (2)
(15/4)a = 15/2 ---> a = 2
Now substituting the value of a back in equation (1) or (2), we get b = 2
The answer is (E)
