Question

How do you solve the quadratic equation $10{{t}^{2}}-29t=-10?$

Answer 1

Hint: We will transpose the constant term from the right-hand side to the left-hand side. Then, we will find the value of $t$ by splitting the middle term. Then we will take the common factors out. Then we will find the value of $t.$

Complete step by step solution:
Let us consider the given polynomial equation in $t, 10{{t}^{2}}-29t=-10.$
Let us transpose $10$ from the right-hand side to the left-hand side of the polynomial equation so that it will become a quadratic equation in $t.$
We will get $10{{t}^{2}}-29t+10=0.$
We need to solve the above quadratic equation to find the values of $t$ that satisfy the quadratic equation.
For, that we will use the sum-product pattern.
We will write the second term as a sum of two numbers. We will get $10{{t}^{2}}-4t-25t+10=0.$
We will take $2t$ out of the first two terms and $-5$ out of the last two terms.
Then, we can write the quadratic equation as $2t\left( 5t-2 \right)-5\left( 5t-2 \right)=0.$
Now, we can take the common factor out.
As a result of taking the common term out, we will get the product $\left( 2t-5 \right)\left( 5t-2 \right)=0.$
Now, we know that this is true only if either $2t-5=0$ or $5t-2=0.$
We know that this is only possible if either $2t=5$ or $5t=2.$
We will transpose the coefficients from the left-hand side to the right-hand side in order to find the value of $t.$
We will get $t=\dfrac{5}{2}$ or $t=\dfrac{2}{5}.$
Hence the solution of the given polynomial equation is $t=\dfrac{5}{2}$ or $t=\dfrac{2}{5}.$

Note: We can also use the quadratic formula to find the solution of the given quadratic equation. We know that the quadratic formula for a polynomial $f\left( x \right)=a{{x}^{2}}+bx+c$ is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.$