Question

find the zeros of 7x^2-18x+8​

Answer 1

Answer:

x = 2 ; x = 4/7

Step-by-step explanation:

$7x {}^{2} - 18x + 8 \\x = \frac{ - b + | - \sqrt{b {}^{2} - 4ac} }{2a} \\ x = \frac{18 +| - \sqrt{324 - 4 \times 7 \times 8} }{2 \times 7}$

$x = \frac{18 +| - \sqrt{324 - 224} }{14} \\ x = \frac{18 + | - \sqrt{100} }{14} \\ x = \frac{18 + | - 10 }{14}$

$x = \frac{18 + 10}{14} = \frac{28}{14} = 2 \\ or \\ x = \frac{18 - 10}{14} = \frac{8}{14} = \frac{4}{7}$

Therefore x can take 2 values;

x = 2 ; x = 4/7

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1. This problem could be solved both by 'quadratic formula' and 'factorisation' method.
2. Here, a = 7

b = -18

c = 8

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Hope it helps.....!

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