To solve the quadratic equation, we make \(y = 0\) and since the question already gave us the quadratic in its factorised form, we will use this to solve to find values for \(x\).

A mathematician has derived an easier way to solve quadratic equation problems, according to MIT's Technology Review. You love challenging math problems. So do we. Let's solve them together ...

Mathematician discovers a simpler way to solve the quadratic equation that could fundamentally change the way students are taught math. A professor at Carnegie Mellon found a new way to solve the ...

Many former algebra students have painful memories of struggling to memorize the quadratic formula. A new way to derive it, overlooked for 4,000 years, is so simple it eliminates the need.

Find A, B, and C for the quadratic equation. If necessary, divide by A so that A now is 1. Write -(B/2) 2-z 2 =C; Solve for z (since you know B and C) Roots are -B/2±z; Everything Old is New Again.

Save guides, add subjects and pick up where you left off with your BBC account. Therefore the coordinates of the turning point are (-1, 2). If we recall the general equation: \(y = a{x^2} + bx + c ...

The prescribed topics to be covered under Quadratic equations are: Standard form of a quadratic equation ax 2 + bx + c = 0, (a ≠ 0). Solutions of quadratic equations (only real roots) by ...

The two solutions to the quadratic equation will be the axis of symmetry plus or minus an unknown amount, which we’ll call u. In this example: r = 2 – u and s = 2 + u .

Again putting the value of a and b in equation (3) we get the value of c. So, we have exactly one choice for each of a,b,c. Hence, the correct option is (C).