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2 years ago

Solve the following equation (x-4)^2=7

2 answers:

7 0

$(x-4)^2 =7\\\\\implies x-4 = \pm \sqrt 7\\\\\implies x = 4 \pm \sqrt 7\\\\\ \text{Hence,}~ x =4 + \sqrt 7~~ \text{or}~~x =4 -\sqrt 7$

melomori [17]2 years ago

5 0

$(x-4) ^ 2 = 7 => | x-4 | = \sqrt{7} \\=> x-4=\sqrt{7} / or/ x-4=-\sqrt{7}\\ => x=\sqrt{7} + 4 /or/ x = 4-\sqrt{7}$

ok done. Thank to me :>

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2 years ago

What is the solution set of the equation using the quadratic formula? x2+6x+10=0 {−3+2i,−3−2i} {−6+2i,−6−2i} {−3+i,−3−i} {−2i,−4

olga2289 [7]

Answer:

The solution set of the quadratic function $x^{2}+6\cdot x +10$ is $\{-3+i,-3-i\}$ .

Step-by-step explanation:

Let be a second-order polynomial (quadratic function) is standard form and equalized to zero:

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Its roots can be determined by the Quadratic Formula in terms of its polynomial coefficients, which states that:

$x_{1,2} = \frac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

Given that $a = 1$ , $b = 6$ and $c = 10$ , the roots of the polynomial are, respectively:

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$x_{1} = -3+i$

$x_{2} = -3 -i$

The solution set of the quadratic function $x^{2}+6\cdot x +10$ is $\{-3+i,-3-i\}$ .

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3 years ago

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