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9p^{2}+27p+14=0
Divide both sides by 2.
a+b=27 ab=9\times 14=126
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9p^{2}+ap+bp+14. To find a and b, set up a system to be solved.
1,126 2,63 3,42 6,21 7,18 9,14
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 126.
1+126=127 2+63=65 3+42=45 6+21=27 7+18=25 9+14=23
Calculate the sum for each pair.
a=6 b=21
The solution is the pair that gives sum 27.
\left(9p^{2}+6p\right)+\left(21p+14\right)
Rewrite 9p^{2}+27p+14 as \left(9p^{2}+6p\right)+\left(21p+14\right).
3p\left(3p+2\right)+7\left(3p+2\right)
Factor out 3p in the first and 7 in the second group.
\left(3p+2\right)\left(3p+7\right)
Factor out common term 3p+2 by using distributive property.
p=-\frac{2}{3} p=-\frac{7}{3}
To find equation solutions, solve 3p+2=0 and 3p+7=0.
18p^{2}+54p+28=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{-54±\sqrt{54^{2}-4\times 18\times 28}}{2\times 18}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 54 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
p=\frac{-54±\sqrt{2916-4\times 18\times 28}}{2\times 18}
Square 54.
p=\frac{-54±\sqrt{2916-72\times 28}}{2\times 18}
Multiply -4 times 18.
p=\frac{-54±\sqrt{2916-2016}}{2\times 18}
Multiply -72 times 28.
p=\frac{-54±\sqrt{900}}{2\times 18}
Add 2916 to -2016.
p=\frac{-54±30}{2\times 18}
Take the square root of 900.
p=\frac{-54±30}{36}
Multiply 2 times 18.
p=-\frac{24}{36}
Now solve the equation p=\frac{-54±30}{36} when ± is plus. Add -54 to 30.
p=-\frac{2}{3}
Reduce the fraction \frac{-24}{36} to lowest terms by extracting and canceling out 12.
p=-\frac{84}{36}
Now solve the equation p=\frac{-54±30}{36} when ± is minus. Subtract 30 from -54.
p=-\frac{7}{3}
Reduce the fraction \frac{-84}{36} to lowest terms by extracting and canceling out 12.
p=-\frac{2}{3} p=-\frac{7}{3}
The equation is now solved.
18p^{2}+54p+28=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.
18p^{2}+54p+28-28=-28
Subtract 28 from both sides of the equation.
18p^{2}+54p=-28
Subtracting 28 from itself leaves 0.
\frac{18p^{2}+54p}{18}=-\frac{28}{18}
Divide both sides by 18.
p^{2}+\frac{54}{18}p=-\frac{28}{18}
Dividing by 18 undoes the multiplication by 18.
p^{2}+3p=-\frac{28}{18}
Divide 54 by 18.
p^{2}+3p=-\frac{14}{9}
Reduce the fraction \frac{-28}{18} to lowest terms by extracting and canceling out 2.
p^{2}+3p+\left(\frac{3}{2}\right)^{2}=-\frac{14}{9}+\left(\frac{3}{2}\right)^{2}
Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
p^{2}+3p+\frac{9}{4}=-\frac{14}{9}+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
p^{2}+3p+\frac{9}{4}=\frac{25}{36}
Add -\frac{14}{9} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(p+\frac{3}{2}\right)^{2}=\frac{25}{36}
Factor p^{2}+3p+\frac{9}{4}. 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+\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{36}}
Take the square root of both sides of the equation.
p+\frac{3}{2}=\frac{5}{6} p+\frac{3}{2}=-\frac{5}{6}
Simplify.
p=-\frac{2}{3} p=-\frac{7}{3}
Subtract \frac{3}{2} from both sides of the equation.
x ^ 2 +3x +\frac{14}{9} = 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.This is achieved by dividing both sides of the equation by 18
r + s = -3 rs = \frac{14}{9}
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{3}{2} - u s = -\frac{3}{2} + u
Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{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='https://opalmath-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-\frac{3}{2} - u) (-\frac{3}{2} + u) = \frac{14}{9}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{14}{9}
\frac{9}{4} - u^2 = \frac{14}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{14}{9}-\frac{9}{4} = -\frac{25}{36}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = \frac{25}{36} u = \pm\sqrt{\frac{25}{36}} = \pm \frac{5}{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{3}{2} - \frac{5}{6} = -2.333 s = -\frac{3}{2} + \frac{5}{6} = -0.667
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.