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Algebra Level 2

If A x 2 + B x + k Ax^2+Bx+k is a perfect square, is it true that k = B 2 4 A k=\frac{ B^2}{4A} ?

True False

James Guevara by James Guevara , 22, Philippines

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2 solutions

Let the expression is S = A x 2 + B x + k Ax^2+Bx+k

We get B = ± 4 A K \pm \sqrt{4AK} = ± 2 A K \pm 2\sqrt{AK} .

S = ( A x ) 2 (\sqrt{A}x)^{2} + 2. A x . K \sqrt{A}x.\sqrt{K} + ( k ) 2 (\sqrt{k})^{2}

= ( ( A ) x ± ( k ) ) 2 (\sqrt(A)x \pm \sqrt(k))^{2}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
James Guevara
Mar 14, 2016

B = 2 √ A k B=2√Ak

B 2 = 4 A k B^2=4Ak

B 2 4 A \frac{B^2}{4A} = 4 A k 4 A \frac{4Ak}{4A}

k= B 2 4 A \frac{B^2}{4A}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you get B=2 root AK?

Anurag Bisht - 3 years, 3 months ago

Let Ax^2+Bx+k = 0 => B^2 - 4Ak = 0

E Koh - 2 years, 2 months ago

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