Quadratic polynomial can be factored using the transformation ax^2+bx+c=aleft(x-x_1 ight)left(x-x_2 ight), wherein x_1 and x_2 space the solutions of the quadratic equation ax^2+bx+c=0.

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every equations the the form ax^2+bx+c=0 can be solved using the quadratic formula: frac-b±sqrtb^2-4ac2a. The quadratic formula gives two solutions, one as soon as ± is addition and one once it is subtraction.
variable the initial expression using ax^2+bx+c=aleft(x-x_1 ight)left(x-x_2 ight). Instead of frac5+sqrt174 for x_1 and also frac5-sqrt174 because that x_2. just how do you differentiate displaystylefleft(x ight)=left(3x+2 ight)left(2x^2-5x+1 ight) utilizing the product rule?
ns founddisplaystylef'left(x ight)=18x^2-22x-7 Explanation:Let us think about the Product Rule:apply it come our case: displaystylef'left(x ight)=3left(2x^2-5x+1 ight)+left(3x+2 ight)left(4x-5 ight)= ...
(x+2)2=2x2-5x+22 Two remedies were discovered : x = 6 x = 3 Rearrange: Rearrange the equation by subtracting what is to the best of the equal sign from both political parties of the equation : ...
3x2-50x+150=0 Two solutions were found : x =(50-√700)/6=(25-5√ 7 )/3= 3.924 x =(50+√700)/6=(25+5√ 7 )/3= 12.743 action by action solution : step 1 :Equation in ~ the end of step 1 : (3x2 - 50x) ...
x2-(-52)x+147=0 Two options were uncovered : x = -3 x = -49 action by action solution : step 1 :Trying to variable by dividing the center term 1.1 Factoring x2+52x+147 The very first term is, ...
2x2-25x+75=0 Two options were found : x = 5 x = 15/2 = 7.500 action by step solution : step 1 :Equation in ~ the finish of step 1 : (2x2 - 25x) + 75 = 0 action 2 :Trying to variable by splitting ...
4x2-15x+14=0 Two remedies were found : x = 7/4 = 1.750 x = 2 step by action solution : step 1 :Equation at the end of action 1 : (22x2 - 15x) + 14 = 0 step 2 :Trying to factor by dividing ...
More Items     Quadratic polynomial can be factored making use of the revolution ax^2+bx+c=aleft(x-x_1 ight)left(x-x_2 ight), whereby x_1 and x_2 room the services of the quadratic equation ax^2+bx+c=0.
All equations of the type ax^2+bx+c=0 can be fixed using the quadratic formula: frac-b±sqrtb^2-4ac2a. The quadratic formula offers two solutions, one as soon as ± is enhancement and one as soon as it is subtraction.

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Factor the initial expression making use of ax^2+bx+c=aleft(x-x_1 ight)left(x-x_2 ight). Substitute frac5+sqrt174 because that x_1 and also frac5-sqrt174 because that x_2.
left< eginarray l l 2 & 3 \ 5 & 4 endarray ight> left< eginarray together l l 2 & 0 & 3 \ -1 & 1 & 5 endarray ight>    EnglishDeutschEspañolFrançaisItalianoPortuguêsРусский简体中文繁體中文Bahasa MelayuBahasa Indonesiaالعربية日本語TürkçePolskiעבריתČeštinaNederlandsMagyar Nyelv한국어SlovenčinaไทยελληνικάRomânăTiếng Việtहिन्दीঅসমীয়াবাংলাગુજરાતીಕನ್ನಡकोंकणीമലയാളംमराठीଓଡ଼ିଆਪੰਜਾਬੀதமிழ்తెలుగు