Determine the sign of $169\cdot7-49\sqrt{123}$

Question

Determine the sign of the discriminant of the equation $$49x^2-26\sqrt7x+\sqrt{123}=0.$$ The coefficients $a,b$ and $c$ of the equation are: $$a=49,\\b=-26\sqrt7\Rightarrow k=-13\sqrt7,\\c=\sqrt{123}.$$ So we have for the discriminant $$D=k^2-ac=(-13\sqrt7)^2-49\cdot\sqrt{123}=169\cdot7-49\sqrt{123}.$$ It is not straight-forward for me which is greater so I tried to compare them $$169\cdot7\mathrel{\Diamond}49\sqrt{123}\\169^2\cdot7^2\mathrel{\Diamond}49^2\cdot123.$$ It is still not obvious. Am I supposed to do the calculations?

Answer 1

Divide $7^2$ from both sides: $$169^2 \leftrightarrow 7^2 \cdot 123 $$ Now $$7^2 \cdot 123 \lt 7^2 \cdot 144 = (7\cdot 12)^2 = 84^2 \lt 169^2 $$ The discriminant is positive.

Answer 2

A different way is to note that $169=13^2$, so $$ \begin{split} 167 \cdot 7 &\lessgtr 49 \sqrt{123}\\ \frac{169}{49} &\lessgtr \sqrt{\frac{123}{49}} = \sqrt{\frac{144}{49} - \frac{21}{49}}\\ \left(\frac{13}{7}\right)^2 &\lessgtr \sqrt{\left(\frac{12}{7}\right)^2 - \frac{21}{49}} \\ \end{split} $$ Since $13/7>1$ the LHS is much larger.

Answer 3

Observe that:

$$\sqrt 7>2~ \text{and} ~ \sqrt {123}<12$$

Therefore,

$$\Delta>(13×2)^2-49×12>0.$$

Answer 4

Keeping the arithmetic and comparisons simple, we have

$$7\sqrt{123}\lt7\sqrt{144}=7\cdot12=84\lt169$$

so $49\sqrt{123}\lt7\cdot169$

Alternatively, since the OP has clearly established that $13^2=169$, we have

$$169\cdot7-49\sqrt{123}\gt169\cdot7-49\sqrt{169}=13^2\cdot7-7^2\cdot13=13\cdot7(13-7)=13\cdot7\cdot6\gt0$$

Answer 5

In the context of quadratic equations, the zeroes are given by $$ r_1 \ , \ r_2 \ = \ -\frac{b}{2a} \ \pm \frac{\sqrt{D}}{2a} \ \ . $$ The equation can be written as $ \ ax^2 + bx + c \ = \ a · (x - r_1) · ( x - r_2) \ = \ 0 \ \ , $ so for this particular polynomial, we have $ \ r_1 + r_2 \ = \ -\frac{b}{a} \ = \frac{26 \sqrt{7}}{49} \ $ and $ \ r_1 · r_2 \ = \ \frac{c}{a} \ = \frac{\sqrt{123}}{49} \ \ . $

Now $ \ r_1 \ - \ r_2 \ = \ \frac{\sqrt{D}}{a} \ . $ Without a calculator, we can estimate $ \ \sqrt{7} \ $ pretty well by $ \ \sqrt{7} \ \approx \ 2.6 \ $ and $ \ 26^2 \ = \ 676 , $ using, say, $ \ (25+1)^2 \ $ , so $ \ 26 \sqrt{7} \ $ is around $ \ 68 \ $ and $ \ \frac{26 \sqrt{7}}{49} \ \approx \ \frac{68}{50} \ $ is roughly $ \ 1.4 \ . $ We also have that $ \ \sqrt{123} \ $ is a little over $ \ 11 \ $ , making $ \ \frac{\sqrt{123}}{49} \ $ about $ \ \frac{11}{50} \ = \ 0.22 \ . $

At last, we have $ (r_1 \ + \ r_2)^2 \ - \ 4·r_1·r_2 \ \approx \ 1.4^2 \ - \ 4·0.22 \ \approx \ 2 \ - \ 0.9 \ \approx \ 1.1 \ . $ This is equal to $ \ (r_1 \ - \ r_2)^2 \ = \ \frac{D}{a^2} \ = \ \frac{D}{49^2} \ , $ so $ \ D \ $ is about $ \ 1.1 · 49^2 \ , $ slightly less than $ \ 1.1 · 2500 \ = \ 2750 \ $ . (Your $ \ 169·7 \ - \ 49 \sqrt{123} \ $ is a factor of $ \ 4 \ $ smaller than $ \ D \ $ , so this checks out.) The quadratic equation thus has two real zeroes.

These zeroes are then about $ \ \frac{68}{100} \ \pm \ \frac{1.05 \ ·\ 50}{100} \ , $ which means they are both positive (but the Rule of Signs already suggested that). A more precise computation with WolframAlpha gives them as about $ \ 1.2180 \ \ \text{and} \ \ 0.1858 \ \ . $