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aleksklad [387]
3 years ago
7

What are the roots of x^2 - 4x - 7 = 0

Mathematics
1 answer:
lyudmila [28]3 years ago
8 0

For the given equation, the solutions are x=2 \pm\sqrt{11} .

Step-by-step explanation:

Step 1:

For an equation of the form ax^{2} +bx+c=0 the solution is x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}.

Here a is the coefficient of x^{2}, b is the coefficient of x and c is the constant term.

Comparing x^{2} -4x-7=0 with ax^{2} +bx+c=0, we get that a is 1, b is -4 and c is -7.

To get the solution, we substitute the values of a, b, and c in x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}.

Step 2:

Substituting the values, we get

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{-(-4) \pm \sqrt{(-4)^{2}-4(1)(-7)}}{2(1)}.

\frac{-(-4) \pm \sqrt{(-4)^{2}-4(1)(-7)}}{2(1)}= \frac{4 \pm \sqrt{16+28}}{2}.

\frac{4 \pm \sqrt{16+28}}{2} = \frac{4 \pm \sqrt{44}}{2}.

\frac{4 \pm \sqrt{44}}{2} =  \frac{4 \pm 2\sqrt{11}}{2} = 2 \pm\sqrt{11} .

So x=2 \pm\sqrt{11} .

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