Question

√
$\sqrt{2x {}^{2} } + 7x + 5 \sqrt{2 } = 0$
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Answer 1

Correct Question:

Solve $\sf{\sqrt{2} x^2+7x+5\sqrt{2} =0}$

To apply the quadratic formula here, first, we convert two coefficient to integers.

This is done by multiplying both sides by √2

⇒ $\sf{2x^2+7\sqrt{2} x+10=0}$

Two solutions are obtained with the quadratic formula.

⇒ $\sf{x=\dfrac{-7\sqrt{2} \pm\sqrt{98-80} }{4} }$

⇒ $\sf{x=\dfrac{-7\sqrt{2} \pm3\sqrt{2} }{4} }$

Hence, two solutions are

⇒ $\sf{x=-\dfrac{5\sqrt{2} }{2} \:or\:x=-\sqrt{2}}$

Learn more:

1. To avoid miscalculation, it is wiser to eliminate irrational coefficients.

So that we don't have to rationalize it again.

2. The point of intersection of a quadratic graph and x-axis is called zero.

The number of the intersection is related to the discriminant.

D>0 ⇒ 2 intersections.

D=0 ⇒ 1 intersection, so the graph is tangent.

D<0 ⇒ 0 intersection. Since the 2 solutions are imaginary.