Solve v^2+5v-6 | Microsoft Math Solver (2024)



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Polynomial5 problems similar to: v ^ { 2 } + 5 v - 6

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a+b=5 ab=1\left(-6\right)=-6

Factor the expression by grouping. First, the expression needs to be rewritten as v^{2}+av+bv-6. To find a and b, set up a system to be solved.

-1,6 -2,3

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

-1+6=5 -2+3=1

Calculate the sum for each pair.

a=-1 b=6

The solution is the pair that gives sum 5.


Rewrite v^{2}+5v-6 as \left(v^{2}-v\right)+\left(6v-6\right).


Factor out v in the first and 6 in the second group.


Factor out common term v-1 by using distributive property.


Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), 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±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.


Square 5.


Multiply -4 times -6.


Add 25 to 24.


Take the square root of 49.


Now solve the equation v=\frac{-5±7}{2} when ± is plus. Add -5 to 7.


Divide 2 by 2.


Now solve the equation v=\frac{-5±7}{2} when ± is minus. Subtract 7 from -5.


Divide -12 by 2.


Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -6 for x_{2}.


Simplify all the expressions of the form p-\left(-q\right) to p+q.

x ^ 2 +5x -6 = 0

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.

r + s = -5 rs = -6

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

r = -\frac{5}{2} - u s = -\frac{5}{2} + u

Two numbers r and s sum up to -5 exactly when the average of the two numbers is \frac{1}{2}*-5 = -\frac{5}{2}. 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. <div style='padding: 8px'><img src='' style='width: 100%;max-width: 700px' /></div>

(-\frac{5}{2} - u) (-\frac{5}{2} + u) = -6

To solve for unknown quantity u, substitute these in the product equation rs = -6

\frac{25}{4} - u^2 = -6

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = -6-\frac{25}{4} = -\frac{49}{4}

Simplify the expression by subtracting \frac{25}{4} on both sides

u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{2}

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r =-\frac{5}{2} - \frac{7}{2} = -6 s = -\frac{5}{2} + \frac{7}{2} = 1

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

Solve v^2+5v-6 | Microsoft Math Solver (2024)
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