Solve for h
h=-3
h=-5
Share
Copied to clipboard
0=-\left(16+8h+h^{2}\right)+1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4-h\right)^{2}.
0=-16-8h-h^{2}+1
To find the opposite of 16+8h+h^{2}, find the opposite of each term.
0=-15-8h-h^{2}
Add -16 and 1 to get -15.
-15-8h-h^{2}=0
Swap sides so that all variable terms are on the left hand side.
-h^{2}-8h-15=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-8 ab=-\left(-15\right)=15
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -h^{2}+ah+bh-15. To find a and b, set up a system to be solved.
-1,-15 -3,-5
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 15.
-1-15=-16 -3-5=-8
Calculate the sum for each pair.
a=-3 b=-5
The solution is the pair that gives sum -8.
\left(-h^{2}-3h\right)+\left(-5h-15\right)
Rewrite -h^{2}-8h-15 as \left(-h^{2}-3h\right)+\left(-5h-15\right).
h\left(-h-3\right)+5\left(-h-3\right)
Factor out h in the first and 5 in the second group.
\left(-h-3\right)\left(h+5\right)
Factor out common term -h-3 by using distributive property.
h=-3 h=-5
To find equation solutions, solve -h-3=0 and h+5=0.
0=-\left(16+8h+h^{2}\right)+1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4-h\right)^{2}.
0=-16-8h-h^{2}+1
To find the opposite of 16+8h+h^{2}, find the opposite of each term.
0=-15-8h-h^{2}
Add -16 and 1 to get -15.
-15-8h-h^{2}=0
Swap sides so that all variable terms are on the left hand side.
-h^{2}-8h-15=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.
h=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -8 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
h=\frac{-\left(-8\right)±\sqrt{64-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}
Square -8.
h=\frac{-\left(-8\right)±\sqrt{64+4\left(-15\right)}}{2\left(-1\right)}
Multiply -4 times -1.
h=\frac{-\left(-8\right)±\sqrt{64-60}}{2\left(-1\right)}
Multiply 4 times -15.
h=\frac{-\left(-8\right)±\sqrt{4}}{2\left(-1\right)}
Add 64 to -60.
h=\frac{-\left(-8\right)±2}{2\left(-1\right)}
Take the square root of 4.
h=\frac{8±2}{2\left(-1\right)}
The opposite of -8 is 8.
h=\frac{8±2}{-2}
Multiply 2 times -1.
h=\frac{10}{-2}
Now solve the equation h=\frac{8±2}{-2} when ± is plus. Add 8 to 2.
h=-5
Divide 10 by -2.
h=\frac{6}{-2}
Now solve the equation h=\frac{8±2}{-2} when ± is minus. Subtract 2 from 8.
h=-3
Divide 6 by -2.
h=-5 h=-3
The equation is now solved.
0=-\left(16+8h+h^{2}\right)+1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4-h\right)^{2}.
0=-16-8h-h^{2}+1
To find the opposite of 16+8h+h^{2}, find the opposite of each term.
0=-15-8h-h^{2}
Add -16 and 1 to get -15.
-15-8h-h^{2}=0
Swap sides so that all variable terms are on the left hand side.
-8h-h^{2}=15
Add 15 to both sides. Anything plus zero gives itself.
-h^{2}-8h=15
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.
\frac{-h^{2}-8h}{-1}=\frac{15}{-1}
Divide both sides by -1.
h^{2}+\left(-\frac{8}{-1}\right)h=\frac{15}{-1}
Dividing by -1 undoes the multiplication by -1.
h^{2}+8h=\frac{15}{-1}
Divide -8 by -1.
h^{2}+8h=-15
Divide 15 by -1.
h^{2}+8h+4^{2}=-15+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
h^{2}+8h+16=-15+16
Square 4.
h^{2}+8h+16=1
Add -15 to 16.
\left(h+4\right)^{2}=1
Factor h^{2}+8h+16. 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(h+4\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
h+4=1 h+4=-1
Simplify.
h=-3 h=-5
Subtract 4 from both sides of the equation.
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}