Algebra 1 question Medal + Fan
The steps to derive the quadratic formula are shown below: Step 1: ax2 + bx + c = 0 Step 2: x2 + b over ax + c over a = 0 Step 3: x2 + b over a x = - c over a Step 4: x2 +b over a x + square of quantity of b over 2 times a = -c over a + square of quantity of b over 2 times a Step 5: x plus b over 2 times a, all squared = -c over a + square of quantity of b over 2 times a Step 6: x + b over 2 times a = square root of all of negative c over a plus the square of the quantity b over 2 times a Step 7: x = - b over 2 times a ± square root of all of negative c over a plus the square of the quantity b over 4 times a = -b over 2 times a ± square root of the quantity of b squared minus 4 times a times c over 4 times a squared= negative b plus or minus square root of b squared minus 4 times a times c over 2 times a Which of the following statements best describes the first incorrect step?
In Step 4, square of quantity of b over 2 times a instead of b over 2 times a was added on both sides. In Step 5, x plus b over 2 times a, all squared was written instead of x plus b over 4 times a, all squared . In Step 6, -c over a was written instead of c over a inside the root. In Step 7, b over 4 times a, all squared was written instead of square of quantity of b over 2 times a inside the root.
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@surjithayer
@jim_thompson5910
can you post a pic of the whole problem?
sure
I agree with your answer, it should be \(\Large \left(\frac{b}{2a}\right)^2\). Look back at step 6.
exactly....
so D it is?
correct
can you help with another @jim_thompson5910 ?
whats your question
Which of the following tables shows the correct steps to transform x2 + 6x + 8 = 0 into the form (x - p)2 = q? [p and q are integers]
you are correct again, nice work
you add 1 to both sides so you have a perfect square trinomial on the left side to factor
thank you :D
yw
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