Question

solve each equation. check the solutions.

1. $x^{2}-6x + 8 = 0$
2. $b^{2}-7b + 12 = 0$
3. $m^{2}+5m + 6 = 0$
4. $d^{2}+7d + 10 = 0$
5. $y^{2}-2y - 24 = 0$
6. $p^{2}-3p - 18 = 0$

challenge (hint: must be set = 0)

1. $h^{2}+2h = 35$
2. $a^{2}+14a=-45$
3. $n^{2}-36 = 5n$
4. $w^{2}+30 = 11w$

Answer:

1. For equation \(x^{2}-6x + 8=0\):

• Step1: Factor the quadratic equation
• We need to find two numbers that multiply to \(8\) and add up to \(-6\). The numbers are \(-2\) and \(-4\). So, \(x^{2}-6x + 8=(x - 2)(x - 4)=0\).
• Step2: Solve for \(x\)
• Using the zero - product property, if \((x - 2)(x - 4)=0\), then \(x-2 = 0\) or \(x - 4=0\).
• For \(x-2=0\), we get \(x=2\). For \(x - 4=0\), we get \(x=4\).

1. For equation \(b^{2}-7b + 12=0\):

• Step1: Factor the quadratic equation
• We find two numbers that multiply to \(12\) and add up to \(-7\). The numbers are \(-3\) and \(-4\). So, \(b^{2}-7b + 12=(b - 3)(b - 4)=0\).
• Step2: Solve for \(b\)
• Using the zero - product property, if \((b - 3)(b - 4)=0\), then \(b-3 = 0\) or \(b - 4=0\).
• For \(b-3=0\), we get \(b=3\). For \(b - 4=0\), we get \(b=4\).

1. For equation \(m^{2}+5m + 6=0\):

• Step1: Factor the quadratic equation
• We find two numbers that multiply to \(6\) and add up to \(5\). The numbers are \(2\) and \(3\). So, \(m^{2}+5m + 6=(m + 2)(m+3)=0\).
• Step2: Solve for \(m\)
• Using the zero - product property, if \((m + 2)(m + 3)=0\), then \(m+2 = 0\) or \(m + 3=0\).
• For \(m+2=0\), we get \(m=-2\). For \(m + 3=0\), we get \(m=-3\).

1. For equation \(d^{2}+7d + 10=0\):

• Step1: Factor the quadratic equation
• We find two numbers that multiply to \(10\) and add up to \(7\). The numbers are \(2\) and \(5\). So, \(d^{2}+7d + 10=(d + 2)(d + 5)=0\).
• Step2: Solve for \(d\)
• Using the zero - product property, if \((d + 2)(d + 5)=0\), then \(d+2 = 0\) or \(d + 5=0\).
• For \(d+2=0\), we get \(d=-2\). For \(d + 5=0\), we get \(d=-5\).

1. For equation \(y^{2}-2y-24=0\):

• Step1: Factor the quadratic equation
• We find two numbers that multiply to \(-24\) and add up to \(-2\). The numbers are \(4\) and \(-6\). So, \(y^{2}-2y-24=(y - 6)(y + 4)=0\).
• Step2: Solve for \(y\)
• Using the zero - product property, if \((y - 6)(y + 4)=0\), then \(y-6 = 0\) or \(y + 4=0\).
• For \(y-6=0\), we get \(y=6\). For \(y + 4=0\), we get \(y=-4\).

1. For equation \(p^{2}-3p-18=0\):

• Step1: Factor the quadratic equation
• We find two numbers that multiply to \(-18\) and add up to \(-3\). The numbers are \(3\) and \(-6\). So, \(p^{2}-3p-18=(p - 6)(p+3)=0\).
• Step2: Solve for \(p\)
• Using the zero - product property, if \((p - 6)(p + 3)=0\), then \(p-6 = 0\) or \(p + 3=0\).
• For \(p-6=0\), we get \(p=6\). For \(p + 3=0\), we get \(p=-3\).

1. For equation \(h^{2}+2h=35\) (rewrite as \(h^{2}+2h - 35=0\)):

• Step1: Factor the quadratic equation
• We find two numbers that multiply to \(-35\) and add up to \(2\). The numbers are \(7\) and \(-5\). So, \(h^{2}+2h - 35=(h + 7)(h-5)=0\).
• Step2: Solve for \(h\)
• Using the zero - product property, if \((h + 7)(h - 5)=0\), then \(h+7 = 0\) or \(h - 5=0\).
• For \(h+7=0\), we get \(h=-7\). For \(h - 5=0\), we get \(h=5\).

1. For equation \(a^{2}+14a=-45\) (rewrite as \(a^{2}+14a + 45=0\)):

• Step1: Factor the quadratic equation
• We find two numbers that multiply to \(45\) and add up to \(14\). The numbers are \(5\) and \(9\). So, \(a^{2}+14a + 45=(a + 5)(a + 9)=0\).
• Step2: Solve for \(a\)
• Using the zero - product property, if \((a + 5)(a + 9)=0\), then \(a+5 = 0\) or \(a + 9=0\).
• For \(a+5=0\), we get \(a=-5\). For \(a + 9=0\), we get \(a=-9\).

1. For equation \(n^{2}-36 = 5n\) (rewrite as \(n^{2}-5n-36=0\)):

• Step1: Factor the quadratic equation
• We find two numbers that multiply to \(-36\) and add up to \(-5\). The numbers are \(4\) and \(-9\). So, \(n^{2}-5n-36=(n - 9)(n + 4)=0\).
• Step2: Solve for \(n\)
• Using the zero - product property, if \((n - 9)(n + 4)=0\), then \(n-9 = 0\) or \(n + 4=0\).
• For \(n-9=0\), we get \(n=9\). For \(n + 4=0\), we get \(n=-4\).

1. For equation \(w^{2}+30=11w\) (rewrite as \(w^{2}-11w + 30=0\)):

• Step1: Factor the quadratic equation
• We find two numbers that multiply to \(30\) and add up to \(-11\). The numbers are \(5\) and \(6\). So, \(w^{2}-11w + 30=(w - 5)(w - 6)=0\).
• Step2: Solve for \(w\)
• Using the zero - product property, if \((w - 5)(w - 6)=0\), then \(w-5 = 0\) or \(w - 6=0\).
• For \(w-5=0\), we get \(w=5\). For \(w - 6=0\), we get \(w=6\).

1. Answer: \(x = 2,x = 4\)
2. Answer: \(b = 3,b = 4\)
3. Answer: \(m=-2,m=-3\)
4. Answer: \(d=-2,d=-5\)
5. Answer: \(y = 6,y=-4\)
6. Answer: \(p=6,p=-3\)
7. Answer: \(h=-7,h=5\)
8. Answer: \(a=-5,a=-9\)
9. Answer: \(n=9,n=-4\)
10. Answer: \(w=5,w=6\)