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solving Quadratic Equation by using Any method.
3.) x2+5x=-6
4.) (x+1)(x-5)=0
5.) x2+7x+10=0
with solution​


Sagot :

SOLVING QUADRATIC EQUATIONS:

3. x²+5x=-6, I'll use the quadratic formula for this one.

  • x²+5x+6=0
  • [tex]x = \frac{ - b \frac{ + }{} \sqrt{b {}^{2} - 4ac} }{2a} [/tex]
  • [tex]x = \frac{ - 5 \frac{ + }{} \sqrt{5 {}^{2} - 4(1)(6)} }{2(1)} [/tex]
  • [tex]x = \frac{ - 5 \frac{ + }{} \sqrt{25 - 24} }{2} = \frac{ - 5 \frac{ + }{} \sqrt{1} }{2} [/tex]
  • [tex]x = \frac{ - 5 \frac{ + }{}1 }{2} [/tex]
  • [tex]x = \frac{ - 5 + 1}{2} = \frac{ - 4}{2} \\ x = - 2[/tex]
  • [tex]x = \frac{ - 5 - 1}{2} = \frac{ - 6}{2} \\ x = - 3[/tex]

checking: x=-2, x=-3

  • x²+5x=-6
  • (-2)²+5(-2)=-6
  • 4-10=-6
  • -6=-6, true

  • x²+5x=-6
  • (-3)²+5(-3)=-6
  • 9-15=-6
  • -6=-6, true

Therefore, the solutions of the equation x²+5x=-6 are x=-2, x=-3.

4. (x+1)(x-5)=0, I'll use the factoring method for this one.

  • (x+1)(x-5)=0
  • x+1=0, x-5=0
  • x=-1, x=5

checking: x=-1, x=5

  • (x+1)(x-5)=0
  • (-1+1)(-1-5)=0
  • (0)(-6)=0
  • 0=0, true

  • (x+1)(x-5)=0
  • (5+1)(5-5)=0
  • (6)(0)=0
  • 0=0, true

Therefore, the solutions of the equation are x=-1, x=5.

5. x²+7x+10=0, I'll use the factoring method for this one.

  • x²+7x+10=0
  • (x+2)(x+5)=0
  • x+2=0, x+5=0
  • x=-2, x=-5

checking: x=-2, x=-5

  • x²+7x+10=0
  • (-2)²+7(-2)+10=0
  • 4-14+10=0
  • -10+10=0
  • 0=0, true

  • x²+7x+10=0
  • (-5)²+7(-5)+10=0
  • 25-35+10=0
  • -10+10=0
  • 0=0, true

Therefore, the solutions are x=-2, x=-5.

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