Question

If one of the root of quadratic equation kx squared -20x+34=0 is 5-2√2. Find k

Answer 1

K = 2

$5 - 2\sqrt{2 } \: \: is \: an \: root \: of \: k {x}^{2} - 20x + 34$

$k {(5 - 2 \sqrt{2} })^{2} \: - 20(5 - 2 \sqrt{2} ) + 34 = 0$

$k \: = \frac{66 - 40 \sqrt{2} }{33 - 20 \sqrt{2} }$

$k \: = \frac{2(33 - 20 \sqrt{2} )}{33 - 20 \sqrt{2} }$

$k \: = 2$

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Answer 2

The value of k is $k=\frac{66-40\sqrt2}{33-20\sqrt2}$

Step-by-step explanation:

Given : If one of the root of quadratic equation $kx^2-20x+34=0$ is 5-2√2.

To find : The value of k ?

Solution :

Quadratic equation $kx^2-20x+34=0$

One of the root is 5-2√2.

So substitute the value of $x=5-2\sqrt2$

$k(5-2\sqrt2)^2-20(5-2\sqrt2)+34=0$

$k(25+8-20\sqrt2)-100+40\sqrt2+34=0$

$k(33-20\sqrt2)=66-40\sqrt2$

$k=\frac{66-40\sqrt2}{33-20\sqrt2}$

Therefore, the value of k is $k=\frac{66-40\sqrt2}{33-20\sqrt2}$

#Learn more

One of tge roots of Quadratic Equation 2x^2+kx-2=0 is -2,find k

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