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Leona [35]
3 years ago
11

Solve 2x2 + 12x − 14 = 0 by completing the square.

Mathematics
2 answers:
horsena [70]3 years ago
8 0
2x^2 + 12x-14 = 0\\2(x^2+6x-7)=0\ /:2\\x^2+6x+9-16=0\\x^2+6x+9=16\\(x-3)^2=4^2\\x-3=4\ \ \ or\ \ \ x-3=-4\\x=7\ \ \ \ \ \ \ \ \ or\ \ \ x=-1\\\\Ans.\ x=7\ or\ x=-1
andrey2020 [161]3 years ago
3 0
2x^2+12x-14=0\ \ \ \ |divide\ both\ sides\ by\ 2\\\\x^2+6x-7=0\\\\x^2+2x\cdot3=7\ \ \ \ |add\ 3^2\ to\ both\ sides\\\\x^2+2x\cdot3+3^2=7+3^2\\\\(x+3)^2=7+9\\\\(x+3)^2=16\iff x+3=-\sqrt{16}\ or\ x+3=\sqrt{16}\\\\x+3=-4\ or\ x+3=4\\\\x=-4-3\ or\ x=4-3\\\\\boxed{x=-7\ or\ x=1}

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Answer:

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

Step-by-step explanation:

First of all, let simplify and factorize the resulting polynomial:

6\cdot x^{2} \geq 10 + 11\cdot x

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Roots are found by Quadratic Formula:

r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}

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Then, the factorized form of the inequation is:

6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0

By Real Algebra, there are two condition that fulfill the inequation:

a) x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0

x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}

x \geq \frac{5}{2}

b) x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0

x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}

x\leq -\frac{2}{3}

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

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