Simple Algebraic Problem

Algebra Level 3

If the minimum value of

$2x^2-3x+2$

can be expressed in the form $\frac{m}{n}$ , where $m$ and $n$ are coprime, positive integers, find the value of $m+n$ .

The answer is 15.

4 solutions

Prasun Biswas
Feb 9, 2015

This problem can be solved simply by completing the square and then using the fact that $A(x-a)^2 \geq 0~\forall x\in \mathbb{R}$ and constants $a,A$ where $A$ must either be zero or positive and equality holds iff $x=a\lor A=0$ .

The given quadratic can be written as follows using the method of completing the square,

$2\left[\left(x-\frac{3}{4}\right)^2+\frac{7}{16}\right]=2\left(x-\frac{3}{4}\right)^2+\frac{7}{8}$

Using the inequality we talked about, we can see that the given quadratic has the minimum $\dfrac{7}{8}$ which is attained when $x=\dfrac{3}{4}$ .

Since $7$ and $8$ are coprime, we have the required sum $=7+8=\boxed{15}$

P.S - This problem can also be solved by calculus using the first and second derivative tests to find maxima or minima at critical points.

There's a name of this inequality I used but I can't seem to remember it.

Prasun Biswas - 6 years, 4 months ago

Parv Maurya
Jan 25, 2015

it is simple, for a quadratic equation if a>0 ,then it's minimum value is -D/4a

Bill Bell
Oct 22, 2014

Quadratic polynomial is a parabola, which is symmetric. Set it to zero and solve. Although in this case roots are complex it doesn't matter. Take their midpoint which is 3/4 and evaluate the polynomial at that value of x since this midpoint between the roots must also be the on the line of symmetry of the parabola and on the minimum of it too.

What you explained is equivalent to the application of derivatives to find local maxima or minima at critical points! If we consider the given function $2x^2-3x+2=f(x)$ , then we have $f'(x)=4x-3$ . At critical points, the slope of the tangent to the curve is $0$ , so the critical points are those for which $f'(x)=0$ . There is only one such point which is $x=\dfrac{3}{4}$ . Performing the second derivative test confirms that this point is indeed a point of local minima and here it's the absolute minimum. So, $f\left(\dfrac{3}{4}\right)=\dfrac{7}{8}$ is the minimum value of the given quadratic polynomial.

But then again, this is an Algebra problem, so you should try to avoid using calculus to solve it.

Prasun Biswas - 6 years, 4 months ago

Hemang Sarkar
Jul 24, 2014

if a is positive then the minimum value of ax^2 + bx + c is (4ac-b^2)/4a at x= -b/2a

Simple Algebraic Problem

The answer is 15.

4 solutions

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