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

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All equations of 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 when ± is addition and one when it is subtraction.
Factor the original expression using ax^2+bx+c=aleft(x-x_1 ight)left(x-x_2 ight). Substitute frac32+sqrt3 for x_1 and frac32-sqrt3 for x_2.
4x^2-12x-3=4left(x-left(sqrt3+frac32 ight) ight)left(x-left(frac32-sqrt3 ight) ight)
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4x2-12x-9 Final result : 4x2 - 12x - 9 Step by step solution : Step 1 :Equation at the end of step 1 : (22x2 - 12x) - 9 Step 2 :Trying to factor by splitting the middle hatchet ...
displaystylex=frac32pmsqrt3displaystylexapprox-0.23quad extandquadxapprox+3.23 Explanation:Given: displaystyle ext 0=4x^2-12x-3 ...
Given that displaystyle extcosh3x+3 extcoshx=4 extcosh^3x how do you show that displaystylek^3=3k-4 has a root in between displaystyle-3 ...
https://socratic.org/questions/given-that-cosh3x-3coshx-4-cosh-3-x-how-do-you-show-that-k-3-3k-4-has-a-root-in-
The real root isdisplaystylek=-2.195823345445647 Explanation:Makingdisplaystylek=y extcoshleft(x ight)and substituting intodisplaystylek^3=3k-4 ...
4x2-12x=-7 Two solutions were found : x =(12-√32)/8=(3-√ 2 )/2= 0.793 x =(12+√32)/8=(3+√ 2 )/2= 2.207 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign ...
4x2-12x-72 Final result : 4 • (x + 3) • (x - 6) Step by step solution : Step 1 :Equation at the end of step 1 : (22x2 - 12x) - 72 Step 2 : Step 3 :Pulling out like terms : 3.1 Pull out ...
Given displaystylefleft(x ight)=2^x,gleft(x ight)=4 how do you determine the combined function displaystyley=left(f-g ight)left(x ight) and state ...
https://socratic.org/questions/given-f-x-2-x-g-x-4-how-do-you-determine-the-combined-function-y-f-g-x-and-state-1
displaystyley=2^x-4 Explanation:The combined functiondisplaystyley=left(f-g ight)left(x ight)simply means thatdisplaystyley=fleft(x ight)-gleft(x ight) ...
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Quadratic polynomial can be factored using the transformation ax^2+bx+c=aleft(x-x_1 ight)left(x-x_2 ight), where x_1 and x_2 are the solutions of the quadratic equation ax^2+bx+c=0.
All equations of 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 when ± is addition and one when it is subtraction.

See more: Let L Be The Line In R3 That Consists Of All Scalar Multiples Of The Vector


4x^2-12x-3=4left(x-left(sqrt3+frac32 ight) ight)left(x-left(frac32-sqrt3 ight) ight)
Factor the original expression using ax^2+bx+c=aleft(x-x_1 ight)left(x-x_2 ight). Substitute frac32+sqrt3 for x_1 and frac32-sqrt3 for x_2.
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 4
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
Two numbers r and s sum up to 3 exactly when the average of the two numbers is frac12*3 = frac32. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u.
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