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Oksanka [162]
4 years ago
14

Solve showing steps. Use Insert equation to get the symbol and Ctrl with . For exponents x2 + 3x - 4 = 0 9x2 + 12x + 4 =0 (leave

answer as a fraction) x2 - 2x - 3 = 0 x2 - 6x + 9 = 0 2x2 - 4x - 3 = 0 (round to two decimal places)*
Mathematics
1 answer:
ipn [44]4 years ago
5 0

Here all the given equations are in the form of

ax^2+bx+c=0

The value of x can be determined by the perfect square method as follows:

ax^2+bx+c=0

\Rightarrow x^2+\frac{b}{a}+\frac{c}{a}=0

\Rightarrow x^2+2\times\frac{b}{2a}+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}=0

\Rightarrow \left(x+\frac{b}{2a}\right)^2=\left(\frac{b}{2a}\right)^2-\frac{c}{a}

\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\;\cdots(i)

Now, for x^2+3x-4=0

a=1, b=3, c=-4

So, from equation (i)

x=\frac{-3\pm\sqrt{3^2-4\times1\times(-4)}}{2\times1}

\Rightarrow x=\frac{3\pm\sqrt{9+16}}{2}

\Rightarrow x=\frac{3\pm5}{2}

\Rightarrow x=\frac{3+5}{2},\frac{3-5}{2}

\Rightarrow x= 4, -1

Similarly, for 9x^2+12x+4=0

a=9, b=12, c=4

So, from equation (i)

x=\frac{-12\pm\sqrt{12^2-4\times9\times4}}{2\times9}

\Rightarrow x=\frac{-12\pm0}{18}

\Rightarrow x=\frac{-2}{3}

For x^2-2x-3=0

a=1, b=-2, c=-3

x=\frac{-(-2)\pm\sqrt{(-2)^2-4\times1\times(-3)}}{2\times1}

\Rightarrow x=\frac{2\pm\sqrt{16}}{2}

\Rightarrow x=\frac{2\pm4}{2}

\Rightarrow x=\frac{6}{2},\frac{-2}{2}

\Rightarrow x=3.00, -1.00

For x^2-6x+9=0

a=1, b=-6, c=9

x=\frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times9}}{2\times1}

\Rightarrow x=\frac{6\pm0}{2}

\Rightarrow x=3.00

For 2x^2-4x-3=0

a=2, b=-4, c=-3

x=\frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-3)}}{2\times2}

\Rightarrow x=\frac{4\pm\sqrt{40}}{4}

\Rightarrow x=\frac{4\pm2\sqrt{10}}{4}

\Rightarrow x=\frac{4+2\sqrt{10}}{4},\frac{4-2\sqrt{10}}{4}

\Rightarrow x=2.58, -0.58

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