Question

Find the zeros of 7=x^2-8x-3 by completing the square

Answer 1

Using the completing the square method, for the equation, we have the zeros as x = √26 + 4 and x = 4 - √26

How can the zeros be found using completing the square method?

The given equation is presented as follows

$7 = \mathbf{ {x}^{2} - 8x - 3}$

Which, by completing the square, gives;

${x}^{2} - 8x - 3 - 7 = 0$

${x}^{2} - 8x - 10 = 0$

${x}^{2} - 8x + {\left(\frac{8}{2} \right) }^{2} - 10 - {\left(\frac{8}{2} \right) }^{2} = 0$

$\mathbf{{(x - 4)}^{2} } =26$

The zeros of the equation;

$7 = {x}^{2} - 8x - 3$

are;

$\underline{x = \pm \sqrt{26} + 4}$

Learn more about completing the square here:

brainly.com/question/10449635

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