(9a+5)(4b+3)factorise problem
Answers
far you have solved linear equations, which include constant terms—plain numbers—and terms with the variable raised to the first power, x^1=xx
1
=xx, start superscript, 1, end superscript, equals, x.
You may have also solved some quadratic equations, which include the variable raised to the second power, by taking the square root from both sides.
In this lesson, you will learn a new way to solve quadratic equations. Specifically you will learn
how to solve factored equations like (x-1)(x+3)=0(x−1)(x+3)=0left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, equals, 0 and
how to use factorization methods in order to bring other equations ((left parenthesislike x^2-3x-10=0)x
2
−3x−10=0)x, squared, minus, 3, x, minus, 10, equals, 0, right parenthesis to a factored form and solve them.
Solving factored quadratic equations
Suppose we are asked to solve the quadratic equation (x-1)(x+3)=0(x−1)(x+3)=0left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, equals, 0. Why is this a quadratic equation?
This is a product of two expressions that is equal to zero. Note that any xxx value that makes either (x-1)(x−1)left parenthesis, x, minus, 1, right parenthesis or (x+3)(x+3)left parenthesis, x, plus, 3, right parenthesis zero, will make their product zero.
\begin{aligned} (x-1)&(x+3)=0 \\\\ \swarrow\quad&\quad\searrow \\\\ x-1=0\quad&\quad x+3=0 \\\\ x=1\quad&\quad x=-3 \end{aligned}
(x−1)
↙
x−1=0
x=1
(x+3)=0
↘
x+3=0
x=−3
Substituting either x=1x=1x, equals, 1 or x=-3x=−3x, equals, minus, 3 into the equation will result in the true statement 0=00=00, equals, 0, so they are both solutions to the equation.