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x^2 – 10x + 13 = k. If one of the solutions to the equation
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22 Jul 2019, 09:13
22 Jul 2019, 09:13
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\(x^2 – 10x + 13 = k\). If one of the solutions to the equation is \(x = 4\), what is the other solution for \(x\)?
Show: :: OA
6
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Re: x^2 – 10x + 13 = k. If one of the solutions to the equation
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28 Jul 2019, 08:18
28 Jul 2019, 08:18
1
Carcass wrote:
\(x^2 – 10x + 13 = k\). If one of the solutions to the equation is \(x = 4\), what is the other solution for \(x\)?
Show: :: OA
6
GIVEN: x² – 10x + 13 = k
If one solution is x = 4, then we can write: 4² – 10(4) + 13 = k
Simplify: 16 - 40 + 13 = k
Simplify: -11 = k
This means our original equation is: x² – 10x + 13 = -11
To solve this quadratic equation, we'll first set it equal to ZERO.
Add 11 to both sides to get: x² – 10x + 24 = 0
Factor to get: (x - 4)(x - 6) = 0
So, EITHER x = 4 OR x = 6
We were already told that x = 4 is one solution.
So, the second solution is x = 6
Answer: 6
Cheers,
Brent
Re: x^2 10x + 13 = k. If one of the solutions to the equation
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04 May 2022, 00:17
04 May 2022, 00:17
1
Carcass wrote:
\(x^2 – 10x + 13 = k\). If one of the solutions to the equation is \(x = 4\), what is the other solution for \(x\)?
Show: :: OA
6
Since we know one of the solutions is 4 => Meaning in -10x one of the components is -4. And a+b = -10 => b=> -6. Crude way. But worked this time.
