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hoa [83]
3 years ago
12

What are the solution for the quadratic equation y=x^2+11x+24

Mathematics
1 answer:
Fed [463]3 years ago
7 0
<span>Simplifying x2 = 11x + -24 Reorder the terms: x2 = -24 + 11x Solving x2 = -24 + 11x Solving for variable 'x'. Reorder the terms: 24 + -11x + x2 = -24 + 11x + 24 + -11x Reorder the terms: 24 + -11x + x2 = -24 + 24 + 11x + -11x Combine like terms: -24 + 24 = 0 24 + -11x + x2 = 0 + 11x + -11x 24 + -11x + x2 = 11x + -11x Combine like terms: 11x + -11x = 0 24 + -11x + x2 = 0 Factor a trinomial. (3 + -1x)(8 + -1x) = 0 Subproblem 1Set the factor '(3 + -1x)' equal to zero and attempt to solve: Simplifying 3 + -1x = 0 Solving 3 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-3' to each side of the equation. 3 + -3 + -1x = 0 + -3 Combine like terms: 3 + -3 = 0 0 + -1x = 0 + -3 -1x = 0 + -3 Combine like terms: 0 + -3 = -3 -1x = -3 Divide each side by '-1'. x = 3 Simplifying x = 3 Subproblem 2Set the factor '(8 + -1x)' equal to zero and attempt to solve: Simplifying 8 + -1x = 0 Solving 8 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-8' to each side of the equation. 8 + -8 + -1x = 0 + -8 Combine like terms: 8 + -8 = 0 0 + -1x = 0 + -8 -1x = 0 + -8 Combine like terms: 0 + -8 = -8 -1x = -8 Divide each side by '-1'. x = 8 Simplifying x = 8Solutionx = {3, 8}</span>
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2 years ago
Which function has a simplified base of 4RootIndex 3 StartRoot 4 EndRoot? f(x) = 2(RootIndex 3 StartRoot 16 EndRoot) Superscript
PIT_PIT [208]

Answer:

  f(x)=4\sqrt[3]{16}^{2x}

Step-by-step explanation:

We believe you're wanting to find a function with an equivalent base of ...

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The functions you're looking at seem to be ...

  f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x

The third choice seems to be the one you're looking for.

9 0
3 years ago
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