aditi1903 wrote:
clearmountain wrote:
If the graph of y = x2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?
(1) 4b = a2 – 4
(2) b = 0
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer is A;
Also could someone tell me if this is a realistic gmat problem from the official test? it seems too technical to be so.
How is it possible only by A..im unable to understand it..can anybody explain?
A short summary about the quadratic equation \(x^2+ax+b=0\). Using a little bit of algebraic manipulation, it can be written as:
\(x^2+2\cdot{\frac{a}{2}}x+\frac{a^2}{4}+b-\frac{a^2}{4}=0\) or \((x+\frac{a}{2})^2=\frac{a^2}{4}-b\).
Now, it is obvious that this equation will have real roots if and only if \(\frac{a^2}{4}-b=\frac{a^2-4b}{4}\geq{0}\) or \(a^2-4b\geq{0}\).
Taking the square root from both sides, we obtain \(|x+\frac{a}{2}|=|\frac{\sqrt{a^2-4b}}{2}|\), from which we get \(x_1=-\frac{a}{2}-\frac{\sqrt{a^2-4b}}{2}\) and \(x_2=-\frac{a}{2}+\frac{\sqrt{a^2-4b}}{2}\).
From here, we can deduce the sum of the two roots being \(x_1+x_2=-a\) and the product \(x_1x_2=b\). These two are the most frequently used formulas for the given quadratic equation. The sum is easy to get, as the two square roots cancel out. For the product, we use the formula \((m+n)(m-n)=m^2-n^2\):
\((-\frac{a}{2}+\frac{\sqrt{a^2-4b}}{2})(-\frac{a}{2}-\frac{\sqrt{a^2-4b}}{2})=(\frac{a}{2})^2-(\frac{\sqrt{a^2-4b}}{2})^2=\frac{a^2}{4}-\frac{a^2-4b}{4}=b\).
In addition, the difference of the roots can be easily deduced and it is \(\sqrt{a^2-4b}\).
On the GMAT, we seldom use the formula which gives the roots, as it is usually much easier to factorize the quadratic expression and immediately find the roots.
Try out for the equation \(x^2-4x+3=0\). Since \(x^2-4x+3=(x-3)(x-1)\), we can immediately see that the two roots are \(1\) and \(3\).
Check out the above formulas and see that you obtain the same roots, product, sum and difference.
The formula for the roots of the quadratic equation appears in the
OG in the Math Review part for the general form \(ax^2+bx+c=0\).
In the above case we have \(a=1, \,\,b=a \,\,and \,\,c=b\).
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