Question - Solving a Quadratic Equation by Factoring
mathsolving quadratic equationfactoring methodquadratic equation solutionfactoring quadratic equation
Solution:
I see an equation written on a piece of paper: \[ g = 5x^2 - 10x + 5 \]To solve this equation for $$ x $$, first, let's see if we can factor it.A quadratic equation in the form $$ ax^2 + bx + c = 0 $$ can sometimes be factored into the form:\[ (dx + e)(fx + h) = 0 \]The given equation $$ g = 5x^2 - 10x + 5 $$ is not set equal to zero, so we can't solve for $$ x $$ until we do so. Let's assume you meant to solve for $$ x $$ when $$ g = 0 $$. That would give us:\[ 0 = 5x^2 - 10x + 5 \]Now, we clearly see that each term is divisible by 5. Let's factor out a 5:\[ 0 = 5(x^2 - 2x + 1) \]The quadratic inside the parentheses is a perfect square trinomial and factors as:\[ 0 = 5(x - 1)^2 \]Now we can find the value of $$ x $$ that makes the equation true by setting $$ (x - 1)^2 $$ equal to zero:\[ (x - 1)^2 = 0 \]Taking the square root of both sides gives us:\[ x - 1 = 0 \]So the solution for $$ x $$ is:\[ x = 1 \]Thus, the equation $$ 5x^2 - 10x + 5 = 0 $$ has a repeated real solution, which is $$ x = 1 $$.
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