To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^2+ax+bx-5. To find a and b, set up a system to be solved.

You are watching: Factor x^2-6x+5


Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.
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displaystylex=-frac32pmfrac12sqrt19 Explanation: displaystyle extusing the method of extcompleting the squaredisplaystyle• ext the coefficient of the x^2 ext term must be 1 ...
3x2+6x-5=0 Two solutions were found : x =(-6-√96)/6=-1-2/3√ 6 = -2.633 x =(-6+√96)/6=-1+2/3√ 6 = 0.633 Step by step solution : Step 1 :Equation at the end of step 1 : (3x2 + 6x) - 5 = 0 ...
4x2+6x-5=0 Two solutions were found : x =(-6-√116)/8=(-3-√ 29 )/4= -2.096 x =(-6+√116)/8=(-3+√ 29 )/4= 0.596 Step by step solution : Step 1 :Equation at the end of step 1 : (22x2 + 6x) - ...
5x2+6x-5=0 Two solutions were found : x =(-6-√136)/10=(-3-√ 34 )/5= -1.766 x =(-6+√136)/10=(-3+√ 34 )/5= 0.566 Step by step solution : Step 1 :Equation at the end of step 1 : (5x2 + 6x) - ...
displaystylex=-3+sqrt14quad extorquadx=-3-sqrt14 Explanation: displaystylex^2+6x-5=0displaystyleRightarrowx^2+6x+left(9-14 ight)=0 ...
x2+6x-56=0 Two solutions were found : x =(-6-√260)/2=-3-√ 65 = -11.062 x =(-6+√260)/2=-3+√ 65 = 5.062 Step by step solution : Step 1 :Trying to factor by splitting the middle hatchet ...
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To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^2+ax+bx-5. To find a and b, set up a system to be solved.
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.
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.
This equation is in standard form: ax^2+bx+c=0. Substitute -1 for a, 6 for b, and -5 for c in the quadratic formula, frac-b±sqrtb^2-4ac2a.
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^2+bx=c.
Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
Factor x^2-6x+9. In general, when x^2+bx+c is a perfect square, it can always be factored as left(x+fracb2 ight)^2.

See more: Which Of The Following Is A Bond In Which Electrons Are Shared Unequally? ?


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.
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 6 exactly when the average of the two numbers is frac12*6 = 3. 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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