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Question: How do you solve the quadratic equation \(10{{t}^{2}}-29t=-10?\)...

How do you solve the quadratic equation 10t2−29t=−10?10{{t}^{2}}-29t=-10?

Explanation

Solution

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

Complete step by step solution:
Let us consider the given polynomial equation in t,10t2−29t=−10.t, 10{{t}^{2}}-29t=-10.
Let us transpose 1010 from the right-hand side to the left-hand side of the polynomial equation so that it will become a quadratic equation in t.t.
We will get 10t2−29t+10=0.10{{t}^{2}}-29t+10=0.
We need to solve the above quadratic equation to find the values of tt 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 10t2−4t−25t+10=0.10{{t}^{2}}-4t-25t+10=0.
We will take 2t2t out of the first two terms and −5-5 out of the last two terms.
Then, we can write the quadratic equation as 2t(5t−2)−5(5t−2)=0.2t\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 (2t−5)(5t−2)=0.\left( 2t-5 \right)\left( 5t-2 \right)=0.
Now, we know that this is true only if either 2t−5=02t-5=0 or 5t−2=0.5t-2=0.
We know that this is only possible if either 2t=52t=5 or 5t=2.5t=2.
We will transpose the coefficients from the left-hand side to the right-hand side in order to find the value of t.t.
We will get t=52t=\dfrac{5}{2} or t=25.t=\dfrac{2}{5}.
Hence the solution of the given polynomial equation is t=52t=\dfrac{5}{2} or t=25.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(x)=ax2+bx+cf\left( x \right)=a{{x}^{2}}+bx+c is given by x=−b±b2−4ac2a.x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.