Question

Find the zeroes of the polynomial p(x) = root2x^2 - 3x - 2root2 .

Answer 1

√2x²-3x-2√2=0
√2x²-4x+x-2√2=0
√2x(x-2√2)+1(x-2√2)=0
(√2x+1)(x-2√2)=0
√2x+1=0
x=-1/√2

x-2√2=0
x=2√2

therefore x=-1/√2,2√2

Answer 2

Answer:

The required zeroes of the polynomial $p(x)=2\sqrt{2},-\frac{1}{\sqrt{2} }$

Step-by-step explanation:

Concept:

Algebraic expressions called polynomials include coefficients and variables. Indeterminates are another name for variables. For polynomial expressions, we can do mathematical operations like addition, subtraction, multiplication, and positive integer exponents, but not division by variable.

Given:

The expression is $p(x)=\sqrt{2} x^{2} -3x-2\sqrt{2}$

To find:

The objective is to find out the zeroes of the polynomial $p(x)$.

Solution:

The given expression is $p(x)=\sqrt{2} x^{2} -3x-2\sqrt{2}$

⇒$p(x)=\sqrt{2} x^{2} -4x+x-2\sqrt{2}$

⇒$p(x)=\sqrt{2}x(x-2\sqrt{2} )+1(x-2\sqrt{2} )$

⇒$p(x)=(x-2\sqrt{2} )(\sqrt{2} x+1)$

$(x-2\sqrt{2} )=0\\x=2\sqrt{2}$

And,

$(\sqrt{2} x+1)=0\\\sqrt{2} x=-1\\x=-\frac{1}{\sqrt{2} }$

Therefore, the required zeroes of the polynomial $p(x)=2\sqrt{2} ,-\frac{1}{\sqrt{2} }$

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