Solve the following ? Please

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Solve the following ? Please

$∣ ∣ x^{2} - 4 x + 3 ∣ ∣ = x + 1$ $|x^2-4x+3|=x+1$

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Answer 1 · 2018-06-17T06:12:39

$x = \frac{5 \pm \sqrt{17}}{2}$ $x=(5+-sqrt17)/2$

Explanation:

If $∣ ∣ x^{2} - 4 x + 3 ∣ ∣ = x + 1$ $|x^2-4x+3|=x+1$ , we have $x \geq - 1$ $x>=-1$ as $∣ ∣ x^{2} - 4 x + 3 ∣ ∣ \geq 0$ $|x^2-4x+3|>=0$ . This gives us the domain of $x$ $x$ .

Now we have two possibilities

$x^{2} - 4 x + 3 = x + 1$ $x^2-4x+3=x+1$ or $x^{2} - 5 x + 2 = 0$ $x^2-5x+2=0$ i.e.

$x = \frac{5 \pm \sqrt{5^{2} - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}$ $x=(5+-sqrt(5^2-8))/2=(5+-sqrt17)/2$

Observe that both are in domain.

If $x^{2} - 4 x + 3 = - x - 1$ $x^2-4x+3=-x-1$ or $x^{2} - 3 x + 4 = 0$ $x^2-3x+4=0$ . But here discriminant is negative $3^{2} - 4 \times 1 \times 4 = - 7$ $3^2-4×1×4=-7$ . Hence we do not have any solution.

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Solve the following ? Please

$∣ ∣ x^{2} - 4 x + 3 ∣ ∣ = x + 1$ $|x^2-4x+3|=x+1$

1 Answer

Explanation:

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Science

Math

Humanities

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Solve the following ? Please

∣∣x2−4x+3∣∣=x+1|x^2-4x+3|=x+1

1 Answer

Explanation:

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$∣ ∣ x^{2} - 4 x + 3 ∣ ∣ = x + 1$ $|x^2-4x+3|=x+1$