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Standard X

Mathematics

Formation of a Quadratic Equation From It's Roots

In each of th...

Question

In each of the following, determine whether the given values are solutions of the given equation or not:

(i) $x^{2} - 3 x + 2 = 0, x = 2, x = - 1$
(ii) $x^{2} + x + 1 = 0, x = 0, x = 1$
(iii) $x^{2} - 3 \sqrt{3} x + 6 = 0, x = \sqrt{3}, x = - 2 \sqrt{3}$
(iv) $x + \frac{1}{x} = \frac{13}{6}, x = \frac{5}{6}, x = \frac{4}{3}$
(v) $2 x^{2} - x + 9 = x^{2} + 4 x + 3, x = 2, x = 3$
(vi) $x^{2} - \sqrt{2} x - 4 = 0, x = - \sqrt{2}, x = - 2 \sqrt{2}$
(vii) $a^{2} x^{2} - 3 a b x + 2 b^{2} = 0, x = \frac{a}{b}, x = \frac{b}{a}$

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Solution

We are given the following quadratic equations and we are asked to find whether the given values are solutions or not

(i)

We have been given that,

Now if is a solution of the equation then it should satisfy the equation

So, substituting in the equation we get

Hence, is a solution of the given quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation

Therefore, from the above results we find out that is a solution and is not a solution of the given quadratic equation.

(ii) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the given quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Therefore, from the above results we find out that both and are not a solution of the given quadratic equation.

(iii) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Therefore, from the above results we find out that and are the solutions of the given quadratic equation.

(iv) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence, is not a solution of the quadratic equation.

Therefore, from the above results we find out that both and are not the solutions of the given quadratic equation.

(v) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the given quadratic equation

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Therefore, from the above results we find out that both and are solutions of the quadratic equation.

(vi) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Therefore, from the above results we find out that is a solution but is not a solution of the given quadratic equation.

(vii) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Therefore, from the above results we find out that is not a solution and is a solution of the given quadratic equation.

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Similar questions

In each of the following determine whether the given values are solution of the given equation or not
$(i) x^{2} - 3 x + 2 = 0, x = 2, x = - 1 (i i) x^{2} + x + 1 = 0, x = 0, x = 1 (i i i) x^{2} - 3 \sqrt{3 x} + 6 = 0, x = \sqrt{3}, x = - 2 \sqrt{3} (i v) x + \frac{1}{x} = \frac{13}{6}, x = \frac{5}{6}, x = \frac{5}{6}, x = \frac{4}{3} (v) 2 x^{2} - x + 9 = x^{2} + 4 x + 3, x = 2, x = 3 (v i) x^{2} - \sqrt{2 x} - 4 = 0, x = - \sqrt{2}, x = - 2 \sqrt{2} (v i i) a^{2} x^{2} - 3 a b x + 2 b^{2} = 0, x = \frac{a}{b}, x = \frac{b}{a}$

In each of the following, determine whether the given values are solutions (roots) of the equation or not :
(i) $3 x^{2} - 2 x - 1$ = 0; x = 1
(ii) $x^{2} + 6 x + 5$ = 0; x = – 1, x = – 5
(iii) $x^{2} + x - 4 = 0$ ; x = – 2

Q. Determine whether the given values are the solutions of the given equation or not.

x^{2} - (\sqrt{2} + \sqrt{3}) x + \sqrt{6} = 0;

x = \sqrt{2}, x = \sqrt{3}

Q. In each of the following, determine whether the given values are solutions of the given equation or not :

x^{2} - 3 \sqrt{3 x} + 6 = 0, x = \sqrt{3}, x = - 2 \sqrt{3}

Q. In each of the following, determine whether the given values are solutions of the given equation or not:

x^{2} - 3 x + 2 = 0, x = 2, x = - 1

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