Solve for p
p=3
p=7
Share
Copied to clipboard
a+b=-10 ab=21
To solve the equation, factor p^{2}-10p+21 using formula p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right). To find a and b, set up a system to be solved.
-1,-21 -3,-7
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 21.
-1-21=-22 -3-7=-10
Calculate the sum for each pair.
a=-7 b=-3
The solution is the pair that gives sum -10.
\left(p-7\right)\left(p-3\right)
Rewrite factored expression \left(p+a\right)\left(p+b\right) using the obtained values.
p=7 p=3
To find equation solutions, solve p-7=0 and p-3=0.
a+b=-10 ab=1\times 21=21
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp+21. To find a and b, set up a system to be solved.
-1,-21 -3,-7
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 21.
-1-21=-22 -3-7=-10
Calculate the sum for each pair.
a=-7 b=-3
The solution is the pair that gives sum -10.
\left(p^{2}-7p\right)+\left(-3p+21\right)
Rewrite p^{2}-10p+21 as \left(p^{2}-7p\right)+\left(-3p+21\right).
p\left(p-7\right)-3\left(p-7\right)
Factor out p in the first and -3 in the second group.
\left(p-7\right)\left(p-3\right)
Factor out common term p-7 by using distributive property.
p=7 p=3
To find equation solutions, solve p-7=0 and p-3=0.
p^{2}-10p+21=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.
p=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 21}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
p=\frac{-\left(-10\right)±\sqrt{100-4\times 21}}{2}
Square -10.
p=\frac{-\left(-10\right)±\sqrt{100-84}}{2}
Multiply -4 times 21.
p=\frac{-\left(-10\right)±\sqrt{16}}{2}
Add 100 to -84.
p=\frac{-\left(-10\right)±4}{2}
Take the square root of 16.
p=\frac{10±4}{2}
The opposite of -10 is 10.
p=\frac{14}{2}
Now solve the equation p=\frac{10±4}{2} when ± is plus. Add 10 to 4.
p=7
Divide 14 by 2.
p=\frac{6}{2}
Now solve the equation p=\frac{10±4}{2} when ± is minus. Subtract 4 from 10.
p=3
Divide 6 by 2.
p=7 p=3
The equation is now solved.
p^{2}-10p+21=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.
p^{2}-10p+21-21=-21
Subtract 21 from both sides of the equation.
p^{2}-10p=-21
Subtracting 21 from itself leaves 0.
p^{2}-10p+\left(-5\right)^{2}=-21+\left(-5\right)^{2}
Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
p^{2}-10p+25=-21+25
Square -5.
p^{2}-10p+25=4
Add -21 to 25.
\left(p-5\right)^{2}=4
Factor p^{2}-10p+25. 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(p-5\right)^{2}}=\sqrt{4}
Take the square root of both sides of the equation.
p-5=2 p-5=-2
Simplify.
p=7 p=3
Add 5 to both sides of the equation.
x ^ 2 -10x +21 = 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 = 10 rs = 21
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 = 5 - u s = 5 + u
Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>
(5 - u) (5 + u) = 21
To solve for unknown quantity u, substitute these in the product equation rs = 21
25 - u^2 = 21
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 21-25 = -4
Simplify the expression by subtracting 25 on both sides
u^2 = 4 u = \pm\sqrt{4} = \pm 2
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =5 - 2 = 3 s = 5 + 2 = 7
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}