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Quiz 16
1.) Factor or state it is prime: $$2x^2-x+13.$$ Solution: Prime
2.) Solve: $$x^2-9=0.$$
Solution: This factors (as a difference of squares): $$x^2-9=(x-3)(x+3)$$ Therefore to solve the equation $x^2-9=0$, we must solve $$(x-3)(x+3)=0.$$ Using the zero product property tells us that the solution is $x=3$ or $x=-3$.
3.) Solve: $$x^2+3x=4.$$ Solution: Subtract $4$ from both sides to get $$x^2+3x-4=0.$$
This factors as $$x^2+3x-4=(x+4)(x-1).$$ Therefore to solve $x^2+3x=4$ we must solve $$(x+4)(x-1)=0.$$ Thus the solution is $x=1$ or $x=-4$. 4.) The product of two consecutive numbers is $156$. Find such a pair of numbers (two pairs of numbers with this property exist).
Solution: Let $x$ be the first of the two numbers. Then $x+1$ is the second. The description says $$x(x+1)=156.$$ This can be rewritten as the quadratic equation $$x^2+x-156=0.$$ Since $12 \cdot (-13)=156$ take $p=12$ and $q=-13$. Factoring by grouping yields $$(x-12)(x+13)=0.$$ Now solving this gives us $x=12$ or $x=-13$. So we can say that two consecutive numbers whose product is $156$ are $12$ and $13$ or $(-13)$ and $(-12)$.
5.) The base of a triangle is $3$ $\mathrm{cm}$ less than its height. Its area is $14$ $\mathrm{cm}^2$. Find the length of the base and the height of this triangle.
Solution: The description says that $b=h-3$. Since the area of any triangle is $A = \dfrac{1}{2}bh$, we plug in $b=h-3$ and $A=28$ to get the equation $$14=\dfrac{1}{2}(h-3)h.$$ Multiplying by $2$ to get rid of the fraction and distributing the $h$ on the right hand side yields $$28=h^2-3h.$$ Subtracting $56$ to put it on the right side yields the equation $$0=h^2-3h-28.$$ We may factor the right hand side and write $$0=(h-7)(h+4).$$ From this we see the solution of the quadratic equation is $h=7$ or $h=-4$. Since $h=-4$ is not physically meaningful, we get the only solution to be $h=7$. So the height of the triangle is $h=7$ and the base is of length $b=7-3=4$.