Question

How do you solve $4{{x}^{2}}=100$ ?

Answer 1

Hint: We are given $4{{x}^{2}}=100$ , we are asked to find the solution of this, to do so we will first understand the type of equation we have, once we get that we reduce the given equation into the standard form after that we will find the greatest common factor from each term then in the remaining term we factorise, using the middle term and lastly we compare it with zero and solve further
We use $a\times c$ in such a way that its sum or difference forms the ‘b’ of the equation $a{{x}^{2}}+bx+c=0$

Complete step-by-step solution:
We are given $4{{x}^{2}}=100$ , we are asked to find the solution to it.
To find the solution of the equation, we should see that as the highest power is 2 so it is or 2 degree polynomial. So it is a quadratic equation.
As our Equation is quadratic equations and we know quadratic equations is given as $a{{x}^{2}}+bx+c=0$ We reduce it to the standard form
So $4{{x}^{2}}=100$ will become
$4{{x}^{2}}-100=0$
Now, to find its solution we will first find the possible greatest common factor of all these.
In 4 and 100 we can see that 4 is the only possible term that can be separated, so we get –
$4\left( {{x}^{2}}-25 \right)=0$
We can write this above equation as
$4\left( {{x}^{2}}+0x-25 \right)=0$
Now we will use the middle term to split
In middle term split apply on $a{{x}^{2}}+bx+c$ , we produce ‘a’ by ‘c’ and then factor ‘ac’ in such a way that
if the product is ‘ac’ while the sum or difference is made up to ‘b’.
Now, we have middle term split on ${{x}^{2}}+0x-25$
We have $a=1$ , $b=0$ and $c=-25$
So,
$a\times c=1\times \left( -25 \right)=-25$
Now we can see that $5\times \left( -5 \right)=-25$ and also $5+\left( -5 \right)=0$
So we use this to split the middle term.
So,
${{x}^{2}}+0x-25={{x}^{2}}+\left( 5-5 \right)x-25$
Opening brackets we get
$={{x}^{2}}+5x-5x-25$
We take common in the first 2 terms and the last 2 terms. So we get –
$=x\left( x+5 \right)-5\left( x+5 \right)$
As $x+5$ is same, so we get –
 $=\left( x-5 \right)\left( x+5 \right)$
So, we get –
${{x}^{2}}-25=\left( x-5 \right)\left( x+5 \right)$
Now we get that –
$4{{x}^{2}}-100=4\left( x-5 \right)\left( x+5 \right)$
So, as $4{{x}^{2}}-100$ . Was equal to zero so we compare Above equation with zero We get
$4\left( x-5 \right)\left( x+5 \right)=0$
Now we use zero product rule which says that $a\times b=0$ means either $a=0$ or $b=0$ So as $4\left( x-5 \right)\left( x+5 \right)=0$
Means either $x+5=0$ or $x-5=0$
Simplifying we get –
$x=-5$ or $x=5$
So we get solution as $x=5$ and -5
Hence for $4{{x}^{2}}=100$ solution are 5 and -5

Note: While find the middle term using factor of $a\times c$ , we need to keep in mind that when the sign of ‘a’ and ‘c’ are same then ‘b’ is obtained by addition only, if the sign of ‘a’ and ‘c’ are different then ‘b’ can be obtained using only subtraction.
So, as we have $a=1$ and $b=-25$ different sign so ‘b’ is obtained as $5-5=0$ using subtraction. Key point to remember is that the degree of the equation will also tell us about the number of solutions. Also 2 degrees means the given equation can have only 2 solutions.