qwertyuioop
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- Jun 14, 2012
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how do you do this
2x^2+5x-7
2x^2+5x-7
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Math
- Thread starter qwertyuioop
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how do you do this
2x^2+5x-7
Do what? Factorize? Cannot be solve the equation for 'x' - you do not have an "=" sign!
Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
Duplicate post
HallsofIvy
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- Jan 27, 2012
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What Subhotosh Kahn and Novice are saying is that you have not yet asked a complete question. What you have so far is exactly like asking "How do I solve this: 'A'? "
Perhaps your question is about solving the equation \(\displaystyle 2x^2+ 5x- 7\)= 0.
If so, there are three pretty standard ways to solve quadratic equations:
1) Try to factor. I use the word "try" because the only reasonable chance to find factors is if we can factor using integer coefficents and that is not always possible. Here, I note the coefficient of \(\displaystyle x^2\) and the constant term are both primes so the only possible integer factors are (a)(x- 7)(2x+ 1) but multiplying it out gives \(\displaystyle 2x^2- 13x- 7\) so that is wrong. (b) (x+ 7)(2x- 1) but multiplying it out gives \(\displaystyle 2x^2+ 13x- 7\) so that is wrong. (c) (x+1)(2x- 7) but multiplying it out gives \(\displaystyle 2x^2- 5x- 7\). (d) \(\displaystyle (x-1)(2x+7)= 2x^2+ 5x- 7\). AHA! And since \(\displaystyle 2x^2+ 5x- 7= (2x+7)(x- 1)= 0\) and the only way two numbers can multiply to give 0 is if one of them is 0- that is, either 2x+7= 0 or x- 1= 0.
2) Complete the square. A "perfect square" is of the form \(\displaystyle (x+ a)^2= x^2+ 2ax+ a^2\). We can first divide both sides by 2 to get \(\displaystyle x^2+ (5/2)x- (7/2)= 0\) and then write it as \(\displaystyle x^2+ (5/2)x= 7/2\). Now compare \(\displaystyle x^2+ (5/2)x\) and \(\displaystyle x^2+ 2ax\). They will be the same if and only if 2a= 5/2 so that a= 5/4. Then \(\displaystyle a^2= 25/16\). Adding 25/16 to both sides of the equation we have \(\displaystyle x^2+ (5/2)x+ 25/16= (x+ 5/4)^2= 7/2+ 25/16= 81/16\). Taking the square root of both sides gives x+ 5/4= 9/4 and x+ 5/4= -9/4.
3) Use the quadratic formula. Applying "completing the square" to the generic "\(\displaystyle ax^2+ bx+ c= 0\)", we arrive at \(\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}\). Here, a= 2, b= 5, and c=-7. Putting those numbers into the formula, \(\displaystyle x= \frac{-5\pm\sqrt{25+ 56}}{4}= \frac{-5\pm\sqrt{81}}{4}= \frac{-5\pm 9}{4}\).
Perhaps your question is about solving the equation \(\displaystyle 2x^2+ 5x- 7\)= 0.
If so, there are three pretty standard ways to solve quadratic equations:
1) Try to factor. I use the word "try" because the only reasonable chance to find factors is if we can factor using integer coefficents and that is not always possible. Here, I note the coefficient of \(\displaystyle x^2\) and the constant term are both primes so the only possible integer factors are (a)(x- 7)(2x+ 1) but multiplying it out gives \(\displaystyle 2x^2- 13x- 7\) so that is wrong. (b) (x+ 7)(2x- 1) but multiplying it out gives \(\displaystyle 2x^2+ 13x- 7\) so that is wrong. (c) (x+1)(2x- 7) but multiplying it out gives \(\displaystyle 2x^2- 5x- 7\). (d) \(\displaystyle (x-1)(2x+7)= 2x^2+ 5x- 7\). AHA! And since \(\displaystyle 2x^2+ 5x- 7= (2x+7)(x- 1)= 0\) and the only way two numbers can multiply to give 0 is if one of them is 0- that is, either 2x+7= 0 or x- 1= 0.
2) Complete the square. A "perfect square" is of the form \(\displaystyle (x+ a)^2= x^2+ 2ax+ a^2\). We can first divide both sides by 2 to get \(\displaystyle x^2+ (5/2)x- (7/2)= 0\) and then write it as \(\displaystyle x^2+ (5/2)x= 7/2\). Now compare \(\displaystyle x^2+ (5/2)x\) and \(\displaystyle x^2+ 2ax\). They will be the same if and only if 2a= 5/2 so that a= 5/4. Then \(\displaystyle a^2= 25/16\). Adding 25/16 to both sides of the equation we have \(\displaystyle x^2+ (5/2)x+ 25/16= (x+ 5/4)^2= 7/2+ 25/16= 81/16\). Taking the square root of both sides gives x+ 5/4= 9/4 and x+ 5/4= -9/4.
3) Use the quadratic formula. Applying "completing the square" to the generic "\(\displaystyle ax^2+ bx+ c= 0\)", we arrive at \(\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}\). Here, a= 2, b= 5, and c=-7. Putting those numbers into the formula, \(\displaystyle x= \frac{-5\pm\sqrt{25+ 56}}{4}= \frac{-5\pm\sqrt{81}}{4}= \frac{-5\pm 9}{4}\).