Question

please solve this.....!!!​

Answer 1

Question:

$x^2 + \sqrt{2x} - 4 = 0; x = \sqrt{2}\:\&\: - 2\sqrt{2}$

Answer:

Here, the two zeros are given. It means that, when we'll substitute the zeros in the place of x the result will be 0.

So, we've to check that, the given zeros of the polynomial is correct or not.

So, substituting x = $\sqrt{2}$ in the given quadratic equation.

$x^2 + \sqrt{2x} - 4 = 0$

$\implies \sqrt{(2)^2} + \sqrt{2} \times \sqrt{2} - 4 = 0$

$\implies 2 + 2 - 4 = 0$

$\implies 4 - 4 = 0$

$\implies 0 = 0$

$\fbox{\bold{\mathsf{L.H.S = R.H.S}}}$

Therefore, $\sqrt{2}$ is a zero of the given quadratic equation.

Substituting x = $- 2\sqrt{2}$ in the given quadratic equation.

$x^2 + \sqrt{2x} - 4 = 0$

$\implies - 2\sqrt{(2) ^2} +(\sqrt{2} \times -2\sqrt{2)} - 4 = 0$

$\implies (4 \times 2) + (-2\sqrt{4}) - 4 = 0$

$\implies 8 + (-2 \times 2) - 4 = 0$

$\implies 8 - 4 - 4 = 0$

$\implies 8 - 8 = 0$

$\implies 0 = 0$

$\fbox{\bold{\mathsf{L.H.S = R.H.S}}}$

Therefore, $- 2\sqrt{2}$ is a zero of the given quadratic equation.