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# Solve for the value of x: $100{x^2} - 20x + 1 = 0$.

Last updated date: 20th Jun 2024
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Hint: We use factorization method to factorize the given equation. We compare the given quadratic equation with the general quadratic equation and write the values of coefficients. Break the coefficient of ‘x’ in such a way that their sum is equal to coefficient of ‘x’ and their product is equal to the product of coefficient of ${x^2}$and the constant term. Equate the factors to zero and calculate the value of x.
* General form of a quadratic equation is $a{x^2} + bx + c = 0$
* Factorization method: If ‘p’ and ‘q’ are the roots of a quadratic equation, then we can say the quadratic equation is $(x - p)(x - q) = 0$

We are given the equation $100{x^2} - 20x + 1 = 0$ … (1)
Since we know general form of a quadratic equation is $a{x^2} + bx + c = 0$
On comparing equation (1) with general equation we can write $a = 100,b = - 20,c = 1$
Now we calculate the product of coefficient of ${x^2}$and the constant term
Since coefficient of ${x^2}$is 100 and the constant term is 1
Then the value of product is $100 \times 1 = 100$
We have to break the coefficient of ‘x’ i.e. -20 in such a way that the sum of those terms gives -20 and the product of those terms gives 100.
We write prime factorization of the number 100
$\Rightarrow 100 = 2 \times 2 \times 5 \times 5$
So we can form $100 = 10 \times 10$
We can write $100 = \left( { - 10} \right) \times \left( { - 10} \right)$ as the product equals 100 and sum equals -20
Substitute the value of$- 20 = \left( { - 10} \right) + \left( { - 10} \right)$ in equation (1)
$\Rightarrow 100{x^2} - 20x + 1 = 100{x^2} - 10x - 10x + 1$
Take ‘10x’ common from first two terms and -1 common from last two terms in RHS
$\Rightarrow 100{x^2} - 20x + 1 = 10x(10x - 1) - 1(10x - 1)$
Take $(10x - 1)$ common and pair the remaining factors in RHS
$\Rightarrow 100{x^2} - 20x + 1 = (10x - 1)(10x - 1)$
Collect powers of same terms using the rule of exponents $a \times a = {a^{1 + 1}} = {a^2}$
$\Rightarrow 100{x^2} - 20x + 1 = {(10x - 1)^2}$
Substitute the value from equation (2) in equation (1)
$\Rightarrow {(10x - 1)^2} = 0$
Take square root on both sides of equation
$\Rightarrow \sqrt {{{(10x - 1)}^2}} = \sqrt 0$
Cancel square root by square power in LHS of the equation
$\Rightarrow 10x - 1 = 0$
Shift constant values to RHS of the equation
$\Rightarrow 10x = 1$
Divide both sides of the equation by 10
$\Rightarrow \dfrac{{10x}}{{10}} = \dfrac{1}{{10}}$
Cancel same factors from numerator and denominator in LHS of the equation
$\Rightarrow x = \dfrac{1}{{10}}$

$\therefore$The value of x is $\dfrac{1}{{10}}$.

Note:
Alternate method:
Another method to find the roots of quadratic equation $100{x^2} - 20x + 1 = 0$
General form of quadratic equation is $a{x^2} + bx + c = 0$ and the roots are given by the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
On comparing the equation $100{x^2} - 20x + 1 = 0$ with general equation
$a = 100,b = - 20,c = 1$
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Put the values
$\Rightarrow x = \dfrac{{ - ( - 20) \pm \sqrt {{{( - 20)}^2} - 4 \times 100 \times 1} }}{{2 \times 100}}$
Write multiplication of two negative signs as positive signs
$\Rightarrow x = \dfrac{{20 \pm \sqrt {400 - 400} }}{{200}}$
$\Rightarrow x = \dfrac{{20}}{{200}}$
Cancel same factors from numerator and denominator
$\Rightarrow x = \dfrac{1}{{10}}$
$\therefore$The value of x is $\dfrac{1}{{10}}$.