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方程x 2-mx m-1 当m为什么数时,关于x的方程（m-2）x^2-mx+2=m-x^2是关于。
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当m为什么数时,关于x的方程（m-2）x^2-mx+2=m-x^2是关于。解：（m-2）x²-mx+2=m-x² 整理，得 （m-1）x²-mx+2-m=0 要使这个方程是关于x的一元二次方程， 则m-1&0，即m&1 所以当m&1时，关于x的方程（m-2）x^2-mx+2=m-x^2是关于x的一元二次方程 这个一元二次方程的二次项系数是m-3， 一次项系数是 -m 常数项是2-m 满意请采纳，不懂请追问，多多好评，呵呵解：（m-2）x²-mx+2=m-x² 整理，得 （m-1）x²-mx+2-m=0 要使这个方程是关于x的一元二次方程， 则m-1&0，即m&1 所以当m&1时，关于x的方程（m-2）x^2-mx+2=m-x^2是关于x的一元二次方程 这个一元二次方程的二次项系数是m-3， 一次项系数是 -m 常数项是2-m 满意请采纳，不懂请追问，多多好评，呵呵二：m-1 一：-m 常：2-m(m-2)x²-mx+2-m+x²=0 (m-1)x²-mx+(2-m)=0 当m&1时，是关于x的一元二次方程 二次项系数m-1 一次项系数：-m 常数项：2-m首先合并同类项 （m-1）x^2-mx+2-m=0 既然是二次方程，所以m-1不等于0,m不等于1 二次项系数（m-1），一次项系数（-m），常数。(m-2)x^2-mx+2=m-x^2整理得（m-1）x^2-mx+2-m=0当二次项系数不是0时，它是关于x的一元二次方程。 所以m不等于1。 二次项系数m-1 。要使这个式子为一元二次方程，就必须有X^2所以m-2&-1即m&1，二次项系数为（m-1）一次项系数为-m，常数项为2-m.m不等于一。二次项系数为负m.常熟项2-m二次项系数（m-1），一次项系数（-m），常数项2-m由题意得，要x有解，即根号b方-4ac有解，得关于m的方程，求m的解是全集 又因为是二次方程，所以m-1不等于0,m不等于1，所以m。已知,四边形ABCD的两边AB,AD的长是关于x的方程x^2-mx+。解：（1） △=b^2-4ac=m^2-4*1*(m/2-1/4)=m^2-2m+1 当AB=AD时，平行四边形ABCD是菱形。此时两根相等，即：△=m^2-2m+1=(m-1)^2=0 ∴m=1时，。。。。.是菱形。 （2）根据求根公式得：X1=(2m-1)/2, X2=1/2. 所以AB=2=X1,AD=1/2 平行四边形ABCD的周长是（2+1/2）*2=5解有x1 = 1/2 , x2 = m-1/21）当x1=x2时，即m=1时，四边形ABCD是菱形，这时菱形的边长=1/22)AB=2即 AD=1/2则周长为5因为是菱形，所以AB=AD所以方程有两个相等的实数根，b2-4ac=o,m=1，将m的值带入原方程，解得x=1/2，边长即为1/2将AB=2带入。已知sina和cosa是方程X^2-MX+M-1=0的两和实数根,且0&a。sina+cosa=Msinacosa=M-1【维达定理】M^2=(sina)^2+2sinacosa+(cosa)^2=1+2【M-1】所以M=1所以sina+cosa=1sinacosa=0 所以a=90&或0&或360&由于0&a&360所以a=90&1)M=12）此时方程为 X^2-+X=0 即X解为1或0∵ 0&a&360则a无解.。已知X=1是关于X的一元二次方程（M2-1）X^2-MX+M^2=0。已知X=1是关于X的一元二次方程（M2-1）X^2-MX+M^2=0的一个根，把x=1代入方程，应该满足得，再求m(m^2-1)-m+m^2=0 2m^2-m-1=0十字相乘法因式分解2m +1 m -1 m -2m(2m+1)(m-1)=0 m=-1/m.或m=1。当m为何值时,关于x方程x^2-mx+m-1=0有两个相等的实数根。根据题意△=m²-4(m-1)=m²-4m+4=(m-2)²=0∴m=2此时方程为：x²-2x+1=0 x1=x2=1解关于x方程x^2-mx+m-1=0有两个相等的实数根即Δ=0即（-m）²-4*1*(m-1)=0即m²-4m+4=0即（m-2）²=0即m=2当m=2时即关于。解：根的判别式△=b^2-4ac=0时，有两个相等的实数根 △=b^2-4ac =(-4)^2-4(m-1/2) =16-4m+2 =18-4m 令△=0 则 18-4m=0 4m=18 m=。△=b^2-4ac =(-4)^2-4(m-1/2) =16-4m+2 =18-4m 令△=0 则 18-4m=0 4m=18 m=4.5将m=4.5代入方程 x^2-4x+4=0 (x-2)^2=0 x=2 答：m=4。上面搞笑吗？错的都复制？的确用判别式，但是算出来m=2，则x=1。关于x的一元二次方程（m-1）x^2-mx+1=0有两个不相等的实。由题意得到△=m²-4(m-1)&0即（m-2）²&0又（m-2）²&0∴m&2因为方程为一元二次方程∴m-1&0即m&1∴﹛m▏m&1,2且m属于R﹜m=1,2 x的值要么就只有一个数，要么是相同的数 除了1.2 其余的值都可以。求证：无论m为何值时,方程x^2-mx+(m-1)=0总有实数根。_。求根公式B^2-4AC 可以知道方程是否有实数根 方程中B=m A=1 C=(m-1) 带入 delta = m^2- 4(m-1)=(m-2)^2 &=0 恒成立 无论m何值，delta恒大于等于0，所以方程只是有一个实数根△=（-m）²-4x1x(m-1)
=m²-4m+4
=(m-2)²&0 所以无论m为何值时，方程x^2-mx+(m-1)=0总有实数根 解判别式为 m^2-4(m-1) =m^2-4m+4 =(m-2)^2 &=0 所以无论m为何值时，方程x^2-mx+(m-1)=0总有实数根。 两根为1和m-1
唉，这题哪用的着判别式，虽然考点是这个，但是不严密。题目没说m一定是实数 所以判别式没用。但是对于这个方程。设关于x的方程（m+1）x^2-mx+m-1=0有实数根时 1.求A集合的时候要分情况： 当二次项系数m+1=0时，x=2满足条件！ 当二次项系数不为0时，由判别式b^2-4ac&0，可解得A={m-2根号3/30，解这个不等式即可，不过，其中由于a与2的大小关系没有确定，所以还得分情况讨论。A&B=B，就是说A包含于B，你通过数轴表示一下就可以了，这样就可以求出a的取值范围。 自己算下吧！希望对你有用！二次函数y=-x²-mx-m+1(x为自变量) 1)x^2-mx-m+1=0有两个解 b^2-4ac= m^2-4(-m+1)=m^2+4m-4&0 m&2根号2-2 xx1 x2&0 x1+x2=m x1*x2=-m+1 y=-m+1 AB=x2-x1=根号[（x1+x2）^2-2x1x2]=根号（m^2+4m-4） AC=根号（x1^2+y^2） BC=根号（x2^2+y^2） △ABC为等腰三角形 如果AC=BC x1^2=x2^2 x1=-x2 （方程有两个实数根，不可能相等） x1+x2=0 m=0 方程x^2+1=0无实数解， AB=AC (x2-x1)^2=x1^2+y^2 x2^2-2x1x2=y^2 x2^2=y^2-2x1x2=(m-1)^2-2(-m+1)=m^2-1 x2=根号（m^2-1） x1=m-根号（m^2-1） x1x2=m根号（m^2-1） -m^2+1=-m+1 m=1 不合题意x2&0 AB=BC (x2-x1)^2=x1^2+y^2 x1^2-2x1x2=y^2 x1^2=m^2-1 x1=根号（m^2-1） 或 -根号（m^2-1） 一样根号（m^2-1）。(1) 若二次函数y=-x²-mx-m+1的图像与x轴有两个交点 则有m^2+4(1-m)=m^2-4m+4=(m-2)^2&0 故m&2的所有实数即为所求m的。当m=______时,关于x的方程x 2-m -mx+1=0是一元一次方程_。解得，∴2-m=0 ∵关于x的方程x 2-m -mx+1=0是一元一次方程，m=2.故答案是
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已知某垄断厂商的短期成本函数为TC＝0.6Q2＋3Q＋2，反需求函数为P＝8－0.4Q，求
已知某垄断厂商的短期成本函数为TC＝0.6Q2＋3Q＋2，反需求函数为P＝8－0.4Q，求
(1)该厂商实现利润最大化时的产量、价格、收益和利润。(2)该厂商实现窢埂促忌讵涣存惟担隶收益最大化时的产量、价格、收益和利润。
(2)该厂商实现收益最大化时的产量、价格、收益和利润。如铁老大。
由STC,解的MC=0.3Q^2-12Q+140.由P=150-5Q得TR=150Q-5Q^2,得M...
(1)由题意可得:MC= 且MR=8-0.8Q 于是,根据利润最大化原则MR=MC有: 8-0.8Q...
(2)该厂商实现收益最大化时的产量、价格、收益和利润。如铁老大。
由反需求函数为P=8-0.4Q得到利润函数曲线为P=8-0.8Q 而单位成本(即供应曲线)为STC/...
由TC可得,MC=1.2Q+3, TR=P*Q=8Q-0.4Q?,推得MR=8-0.8Q MC=MR...
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Is x & 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|Liked the question? encourage by giving kudos
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Is x & 0?(1) \(|x+3|=4x-3\) --& LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --& \(4x-3\geq{0}\) --& \(x\geq{\frac{3}{4}}\), hence \(x&0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --&
\((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --& \(x(3x-6)=0\) --& \(x=0\) or \(x=2\). Not sufficient.Answer: A.Hope it helps.
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Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)a. |x+3| = 4x -3
|x+3| = x + 3 for x + 3 &= 0 ---& x & -3.
Solving for x,
x + 3 = 4x -3 ---& 6 = 3x ---& x = 2 ( solution accepted since x & -3)
| x + 3| =
-(x +3 ) for x + 3 & 0 --& x & -3.
Solving for x,
-x - 3 = 4x -3
5x = 0 ---& x = 0 ( solution discarded as x is not & -3)
A is sufficient.b. |x -3|
|x -3| = x -3 for x-3 &= 0 ---& x &= 3
|x -3| = -(x -3) for x-3 & 0 ---& x & 3
|2x -3 | = 2x -3 for 2x -3 &= 0 ----& x &=3/2
|2x -3 | = -(2x -3) for 2x -3 & 0 ----& x & 3/2So we have 3 ranges 1) x &=3 ==& x - 3 = 2x -3 ====& x
= 0 ; discard2) 3/2 & x & 3 ===& -(x-3) = 2x -3 ====& -x + 3 = 2x -3 ===& x =2 ( accepted solution )3) x & 3/2 ===& -(x-3) = -(2x -3) ====& x = 3 (discarded the solution)Let me know if I have solved the question correctly. I know the process is lengthy. One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case
3/2 & x & 3), then both statements are sufficient independently and we get the answer as option (D). Please discuss.Regards
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saxenaashi wrote:Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)a. |x+3| = 4x -3
|x+3| = x + 3 for x + 3 &= 0 ---& x & -3.
Solving for x,
x + 3 = 4x -3 ---& 6 = 3x ---& x = 2 ( solution accepted since x & -3)
| x + 3| =
-(x +3 ) for x + 3 & 0 --& x & -3.
Solving for x,
-x - 3 = 4x -3
5x = 0 ---& x = 0 ( solution discarded as x is not & -3)
A is sufficient.b. |x -3|
|x -3| = x -3 for x-3 &= 0 ---& x &= 3
|x -3| = -(x -3) for x-3 & 0 ---& x & 3
|2x -3 | = 2x -3 for 2x -3 &= 0 ----& x &=3/2
|2x -3 | = -(2x -3) for 2x -3 & 0 ----& x & 3/2So we have 3 ranges 1) x &=3 ==& x - 3 = 2x -3 ====& x
= 0 ; discard2) 3/2 & x & 3 ===& -(x-3) = 2x -3 ====& -x + 3 = 2x -3 ===& x =2 ( accepted solution )3) x & 3/2 ===& -(x-3) = -(2x -3) ====& x = 3 (discarded the solution)Let me know if I have solved the question correctly. I know the process is lengthy. One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case
3/2 & x & 3), then both statements are sufficient independently and we get the answer as option (D). Please discuss.RegardsIt seems that you understand this method very well. Everything is correct except the red parts: -(x-3) = -(2x -3) --& x=0 (not x=3), so it's also a valid solution. Hence for (2) you have two valid solutions x=2 and x=0 (just like in my post above), which makes this statement not sufficient.Hope it's clear.
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Bunuel wrote:saxenaashi wrote:Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)a. |x+3| = 4x -3
|x+3| = x + 3 for x + 3 &= 0 ---& x & -3.
Solving for x,
x + 3 = 4x -3 ---& 6 = 3x ---& x = 2 ( solution accepted since x & -3)
| x + 3| =
-(x +3 ) for x + 3 & 0 --& x & -3.
Solving for x,
-x - 3 = 4x -3
5x = 0 ---& x = 0 ( solution discarded as x is not & -3)
A is sufficient.b. |x -3|
|x -3| = x -3 for x-3 &= 0 ---& x &= 3
|x -3| = -(x -3) for x-3 & 0 ---& x & 3
|2x -3 | = 2x -3 for 2x -3 &= 0 ----& x &=3/2
|2x -3 | = -(2x -3) for 2x -3 & 0 ----& x & 3/2So we have 3 ranges 1) x &=3 ==& x - 3 = 2x -3 ====& x
= 0 ; discard2) 3/2 & x & 3 ===& -(x-3) = 2x -3 ====& -x + 3 = 2x -3 ===& x =2 ( accepted solution )3) x & 3/2 ===& -(x-3) = -(2x -3) ====& x = 3 (discarded the solution)Let me know if I have solved the question correctly. I know the process is lengthy. One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case
3/2 & x & 3), then both statements are sufficient independently and we get the answer as option (D). Please discuss.RegardsIt seems that you understand this method very well. Everything is correct except the red parts: -(x-3) = -(2x -3) --& x=0 (not x=3), so it's also a valid solution. Hence for (2) you have two valid solutions x=2 and x=0 (just like in my post above), which makes this statement not sufficient.Hope it's clear.Works. Thanks for pointing out the mistake and quick response too. I am just learning this so working ground up, hence I am not hesitant in taking long route initially. After next few questions, I would be tuned to extract the ranges
I guess. Is there a condition on squaring the sides of the equation or inequation.Regards
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saxenaashi wrote:Works. Thanks for pointing out the mistake and quick response too. I am just learning this so working ground up, hence I am not hesitant in taking long route initially. After next few questions, I would be tuned to extract the ranges
I guess. Is there a condition on squaring the sides of the equation or inequation.RegardsA. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).For example: \(2&4\) --& we can square both sides and write: \(2^2&4^2\);\(0\leq{x}&{y}\) --& we can square both sides and write: \(x^2&y^2\);But if either of side is negative then raising to even power doesn't always work.For example: \(1&-2\) if we square we'll get \(1&4\) which is not right. So if given that \(x&y\) then we can not square both sides and write \(x^2&y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: \(-2&-1\) --& we can raise both sides to third power and write: \(-2^3=-8&-1=-1^3\) or \(-5&1\) --& \(-5^2=-125&1=1^3\);\(x&y\) --& we can raise both sides to third power and write: \(x^3&y^3\).So in statement (2) since both parts of expression are non-negative we can safely apply squaring.Hope it helps.
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Well, I have a little bit another solution for this problem Consider (1): if x&-3 then x+3=4x-3, so x=2, sufficient if x&-3, then -x-3=4x-3, which Leeds to answer x=0, which is wrong because we consider only x&-3Sum up (1) is sufficient The saim logic can be applied to the (2), and it is possible to derive that (2) is sufficient So the answer is D either (1) is sufficient or (2) is sufficientIn your logic, you gave a mistake, because you've missed the root. Posted from
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terance wrote:Well, I have a little bit another solution for this problem Consider (1): if x&-3 then x+3=4x-3, so x=2, sufficient if x&-3, then -x-3=4x-3, which Leeds to answer x=0, which is wrong because we consider only x&-3Sum up (1) is sufficient The saim logic can be applied to the (2), and it is possible to derive that (2) is sufficient So the answer is D either (1) is sufficient or (2) is sufficientIn your logic, you gave a mistake, because you've missed the root. Posted from Welcome to GMAT Club.Unfortunately your answer is not correct: OA for this question is A, not D (you can see it under the spoiler in the initial post).(2) |x – 3| = |2x – 3| has two roots x=0 and x=2 (just substitute these values to see that they both satisfy the given equation), so you can not get the single numerical value of x, which makes this statement insufficient.You can refer to above solutions for two different approaches of how to get these roots for (2). Please ask if anything remains unclear.
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A simple solution for this is solve for x in each statement, but remember when you solve for x with abs value you must make the term positive and negative. in both cases you will find x = +# and zero.
plug the numbers back into the equation and you will find zero will not fit the equation so x must be & 0
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dvinoth86 wrote:Is x & 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|Liked the question? encourage by giving kudos
Remember: When you have || on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have || on only one side you MUST VERIFY the answer by putting them back in eqn(1) |x + 3| = 4x – 3
a) x + 3 = 4x – 3 =& x=2 .. VALID b) -(x + 3) = 4x – 3 =& x=0 .. INVALIDHence Sufficient(2) |x – 3| = |2x – 3| =& x = 0 or 6 .. INVALIDHence In-Sufficient
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Is this a correct way to approach 2) |x – 3| = |2x – 3| ?Step 1:(x-3) = (2x-3)=& x = 0Substitute x = 0 in |x – 3| = |2x – 3| =& VALIDStep 2:(x-3) = -(2x-3)=& x = 2Substitute x = 2 in |x – 3| = |2x – 3| =& VALIDSo 2) |x – 3| = |2x – 3| is NOT SUFFICIENT
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kirncha wrote:Is this a correct way to approach 2) |x – 3| = |2x – 3| ?Step 1:(x-3) = (2x-3)=& x = 0Substitute x = 0 in |x – 3| = |2x – 3| =& VALIDStep 2:(x-3) = -(2x-3)=& x = 2Substitute x = 2 in |x – 3| = |2x – 3| =& VALIDSo 2) |x – 3| = |2x – 3| is NOT SUFFICIENTYes but you can skip the substitution part, read my statement above
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Bunuel wrote:Is x & 0?(1) \(|x+3|=4x-3\) --& LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --& \(4x-3\geq{0}\) --& \(x\geq{\frac{3}{4}}\), hence \(x&0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --&
\((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --& \(x(3x-6)=0\) --& \(x=0\) or \(x=2\). Not sufficient.Answer: A.Hope it helps.Pretty slick answer. Saves a lot of time.
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Hi BunuelThis is how I solved the 2 statements, and I get 0 & 2 for both statements, can u tell me what wrong am I doing?Posted from my mobile device
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nikhil007 wrote:Hi BunuelThis is how I solved the 2 statements, and I get 0 & 2 for both statements, can u tell me what wrong am I doing?Posted from my mobile device If you substitute x=0 in |x + 3| = 4x – 3 you'll get: LHS=|x + 3|=3 and RHS=4x – 3=-3, thus \(LHS\neq{RHS}\), which means that x=0 is not the root of the given equation.When expanding |x+3|:When x&-3, then |x+3|=-(x+3), so in this case we'll have -(x+3)=4x-3 --& x=0 --& discard this value since 0 is not less than -3 (we consider the range when x&-3).When x&=-3, then |x+3|=x+3, so in this case we'll have x+3=4x-3 --& x=2 --& this value of x is OK since 2&-3.So, |x + 3| = 4x - 3 has only one root, x=2.Hope it's clear.
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Bunuel wrote:nikhil007 wrote:Hi BunuelThis is how I solved the 2 statements, and I get 0 & 2 for both statements, can u tell me what wrong am I doing?Posted from my mobile device If you substitute x=0 in |x + 3| = 4x – 3 you'll get: LHS=|x + 3|=3 and RHS=4x – 3=-3, thus \(LHS\neq{RHS}\), which means that x=0 is not the root of the given equation.When expanding |x+3|:When x&-3, then |x+3|=-(x+3), so in this case we'll have -(x+3)=4x-3 --& x=0 --& discard this value since 0 is not less than -3 (we consider the range when x&-3).When x&=-3, then |x+3|=x+3, so in this case we'll have x+3=4x-3 --& x=2 --& this value of x is OK since 2&-3.So, |x + 3| = 4x - 3 has only one root, x=2.Hope it's clear.Thanks Bunuel, this is how I solve the Modulus questions, so as a rule i guess, I should check both values by putting them back in eq, I use to think that they both are the roots since I considered both scenarios, ie one with |X+3| and one with -(x+3).I guess I will have to add this additional step in my methodology.Thanks.
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dvinoth86 wrote:Is x & 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|Liked the question? encourage by giving kudos
I generally think using graphs are faster for simple linear equations
but you should be fast at plotting them as soon as you see the equations.For statement 2, if you plot the modulus graphs , you will see that one root is at x=0 and another root is some +ve x, 2 solutions hence insufficienti.e f(x) = x-3 , |f(0)| = 3(y intercept) and x intercept is @ x=3second f(x) = 2x -3 , |f(0)| = 3( y intercept is same as previous graph or solution 1)
and x intercept is 3/2.
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Hi!Why do we not find the positive and negative values for |x+3|=4x-3?
Is it because this isn't a &, &, &= problem?Thanks! Bunuel wrote:Is x & 0?(1) \(|x+3|=4x-3\) --& LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --& \(4x-3\geq{0}\) --& \(x\geq{\frac{3}{4}}\), hence \(x&0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --&
\((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --& \(x(3x-6)=0\) --& \(x=0\) or \(x=2\). Not sufficient.Answer: A.Hope it helps.
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WholeLottaLove wrote:Hi!Why do we not find the positive and negative values for |x+3|=4x-3?
Is it because this isn't a &, &, &= problem?Thanks! Bunuel wrote:Is x & 0?(1) \(|x+3|=4x-3\) --& LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --& \(4x-3\geq{0}\) --& \(x\geq{\frac{3}{4}}\), hence \(x&0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --&
\((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --& \(x(3x-6)=0\) --& \(x=0\) or \(x=2\). Not sufficient.Answer: A.Hope it helps.One can do this way too, but the way shown in my post is faster.
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Hmmm...When I solve for the pos. and neg. values of |x+3|=4x-3 I get:I. x+3=4x-3
x=2II. x+3=-(4x-3)
x=0So here in my presumably incorrect simplification, I have x=2 and x=0 in which case we can't be sure if x&0For many abs. value questions it seems that you have to find the positive and negative cases each equation.
I get that in this case, |x+3|=4x-3 means that 4x-3 is positive but why for other, similar questions, is solving for the positive and negative cases necessary?Bunuel wrote:WholeLottaLove wrote:Hi!Why do we not find the positive and negative values for |x+3|=4x-3?
Is it because this isn't a &, &, &= problem?Thanks! Bunuel wrote:Is x & 0?(1) \(|x+3|=4x-3\) --& LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --& \(4x-3\geq{0}\) --& \(x\geq{\frac{3}{4}}\), hence \(x&0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --&
\((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --& \(x(3x-6)=0\) --& \(x=0\) or \(x=2\). Not sufficient.Answer: A.Hope it helps.One can do this way too, but the way shown in my post is faster.
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