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a+b=6 ab=-40
To solve the equation, factor r^{2}+6r-40 using formula r^{2}+\left(a+b\right)r+ab=\left(r+a\right)\left(r+b\right). To find a and b, set up a system to be solved.
-1,40 -2,20 -4,10 -5,8
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 -40.
-1+40=39 -2+20=18 -4+10=6 -5+8=3
Calculate the sum for each pair.
a=-4 b=10
The solution is the pair that gives sum 6.
\left(r-4\right)\left(r+10\right)
Rewrite factored expression \left(r+a\right)\left(r+b\right) using the obtained values.
r=4 r=-10
To find equation solutions, solve r-4=0 and r+10=0.
a+b=6 ab=1\left(-40\right)=-40
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as r^{2}+ar+br-40. To find a and b, set up a system to be solved.
-1,40 -2,20 -4,10 -5,8
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 -40.
-1+40=39 -2+20=18 -4+10=6 -5+8=3
Calculate the sum for each pair.
a=-4 b=10
The solution is the pair that gives sum 6.
\left(r^{2}-4r\right)+\left(10r-40\right)
Rewrite r^{2}+6r-40 as \left(r^{2}-4r\right)+\left(10r-40\right).
r\left(r-4\right)+10\left(r-4\right)
Factor out r in the first and 10 in the second group.
\left(r-4\right)\left(r+10\right)
Factor out common term r-4 by using distributive property.
r=4 r=-10
To find equation solutions, solve r-4=0 and r+10=0.
r^{2}+6r-40=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.
r=\frac{-6±\sqrt{6^{2}-4\left(-40\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
r=\frac{-6±\sqrt{36-4\left(-40\right)}}{2}
Square 6.
r=\frac{-6±\sqrt{36+160}}{2}
Multiply -4 times -40.
r=\frac{-6±\sqrt{196}}{2}
Add 36 to 160.
r=\frac{-6±14}{2}
Take the square root of 196.
r=\frac{8}{2}
Now solve the equation r=\frac{-6±14}{2} when ± is plus. Add -6 to 14.
r=4
Divide 8 by 2.
r=-\frac{20}{2}
Now solve the equation r=\frac{-6±14}{2} when ± is minus. Subtract 14 from -6.
r=-10
Divide -20 by 2.
r=4 r=-10
The equation is now solved.
r^{2}+6r-40=0
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.
r^{2}+6r-40-\left(-40\right)=-\left(-40\right)
Add 40 to both sides of the equation.
r^{2}+6r=-\left(-40\right)
Subtracting -40 from itself leaves 0.
r^{2}+6r=40
Subtract -40 from 0.
r^{2}+6r+3^{2}=40+3^{2}
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.
r^{2}+6r+9=40+9
Square 3.
r^{2}+6r+9=49
Add 40 to 9.
\left(r+3\right)^{2}=49
Factor r^{2}+6r+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(r+3\right)^{2}}=\sqrt{49}
Take the square root of both sides of the equation.
r+3=7 r+3=-7
Simplify.
r=4 r=-10
Subtract 3 from both sides of the equation.
x ^ 2 +6x -40 = 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 = -6 rs = -40
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 = -3 - u s = -3 + u
Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-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. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-3 - u) (-3 + u) = -40
To solve for unknown quantity u, substitute these in the product equation rs = -40
9 - u^2 = -40
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -40-9 = -49
Simplify the expression by subtracting 9 on both sides
u^2 = 49 u = \pm\sqrt{49} = \pm 7
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-3 - 7 = -10 s = -3 + 7 = 4
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.