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quadratic function in Algebra
please solve the following:

1. f(x)= x2+6x-4
2. x2-676
3. 9x2-30=6
4. x2+8x+7=0
5. x2+4x-3=0
6. x2+2x-1=0

thanks!

Sagot :

riza1
[tex]1.)\\\\ f(x)= x^2+6x-4 \\ \\x^2+6x-4=0\\\\a=1, \ \ b=6, \ \ c=-4[/tex]

[tex]x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-6-\sqrt{6^2-4 \cdot 1\cdot (-4) }}{2 }=\frac{-6-\sqrt{36+16 }}{2 }=\\\\=\frac{-6-\sqrt{52 }}{2 } =\frac{-6-\sqrt{4\cdot 13 }}{2 }=\frac{-6-2\sqrt{ 13 }}{2 }=\frac{2(-3- \sqrt{ 13 })}{2 }=-3- \sqrt{ 13 }\\\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-6+\sqrt{6^2-4 \cdot 1\cdot (-4) }}{2 }=\frac{2(-3+\sqrt{ 13 })}{2 }=-3+ \sqrt{ 13 }[/tex]


[tex]2.)\\\\ x^2-676 =0 \\ \\(x-26)(x+26)=0\\\\x-26=0 \ \ or \ \ x+26 =0 \\\\x=26 \ \ or \ \ x=-26[/tex]


[tex]3.) 9x^2-30=6 \\\\9x^2-30-6=0\\\\ 9x^2-36=0\\\\(3x-6)(3x+6)=0\\\\3x-6=0\ \ \ or \ \ \ 3x+6=0[/tex]

[tex]3x=6 \ \ \ or \ \ \ 3x=-6\ \ / | \ divide \ both \ sides\ by\ 3 \\\\x=2 \ \ \ or \ \ \ x=-2[/tex]


[tex]4.)\\\\ x^2+8x+7=0\\\\ x^2+x+7x+7=0\\\\x(x +1)+7( x+1)=0\\\\(x+1)(x+7)=0[/tex]

[tex]x+1=0 \ \ \ or \ \ \ x+7 =0 \\\\x=-1 \ \ \ or \ \ \ x=-7[/tex]


[tex]5. )x^2+4x-3=0\\\\ a=1, \ \ \ b=4, \ \ \ c=-3[/tex]

[tex]x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-4-\sqrt{4^2-4 \cdot 1\cdot (- 3) }}{2 }=\frac{-4-\sqrt{16+12 }}{2 }=\\\\=\frac{-4-\sqrt{28 }}{2 }=\frac{-4-\sqrt{4\cdot 7 }}{2 }=\frac{-4-2\sqrt{7 }}{2 } =\frac{2(-2- \sqrt{7 })}{2 }=-2-\sqrt{7}\\\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{-4+\sqrt{4^2-4 \cdot 1\cdot (- 3) }}{2 }= \frac{2(-2+ \sqrt{7 })}{2 }=-2+\sqrt{7}[/tex]


[tex]6. )\\\\ x^2+2x-1=0 \\ \\a=1, \ \ \ \b=2, \ \ \ c=-1\\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-2-\sqrt{2^2-4 \cdot 1\cdot (- 1) }}{2 }=\frac{-2-\sqrt{4+ 4 }}{2 }= \frac{-2-\sqrt{8 }}{2 }= \\\\= \frac{-2-\sqrt{4\cdot 2 }}{2 } =\frac{-2-2\sqrt{ 2 }}{2 }=\frac{2(-1- \sqrt{ 2 })}{2 } =-1- \sqrt{ 2 }[/tex]

[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{-2+\sqrt{2^2-4 \cdot 1\cdot (- 1) }}{2 }= \frac{2(-1+ \sqrt{ 2 })}{2 } =-1+\sqrt{ 2 }[/tex]