What are the approximate solutions of 5x2−7x=1 rounded to the nearest hundredth?

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What are the approximate solutions of  5x2−7x=1 rounded to the nearest hundredth?

Devyn Figueroa · Accepted Answer

Step 1 Subtracting 1 from both sides we get: 5x2−7x−1=0 This is of the form ax2+bx+c=0 with a=5, b=−7 and c=−1 The general formula for roots of such a quadratic gives us: x=−b±b2−4ac2a =7±(−7)2−(4×5×−1)2×5 =7±6910 =0.7±6910 What is a good approximation for 69? We could punch it into a calculator, but let&#039;s do it by hand instead using Newton-Raphson: 82=64, so 8 seems like a good first approximation Then iterate using the formula: an+1=a{n}2+692an Let a0=8 a1=64+6916=13316=8.3125 This is almost certainly good enough for the accuracy requested. So 6910=∼8.310=0.83 x=∼0.7±0.83 That is x=∼1.53 or x=∼−0.13

Bottisiooq · Accepted Answer

Step 1  Rewrite 5x2−7x=1 in the standar form of ax2+bx+c=0 giving  5x2−7x−1=0  then use the Quadratic Formula for roots:  x=−b±b2−4ac2a  In this case  x=7pn49+2010  Using a calculator:  69=8.306624 (approx.)  So  x=15.30662410=1.53 (rounded to nearest hundredth)  or  x=−1.30662410=−0.13 (rounded to the nearest hundredth)

What are the approximate solutions of 5x^{2}-7x=1 rounded to the nearest

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