Which term is added in x^2+6 x+7=0 to make it complete square ? (… - Vidyadip

Question

Which term is added in $x^2+6 x+7=0$ to make it complete square ? $(1,2,3)$

✓

Answer

2

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "Fill In The Blanks[1 Marks ]" from Quadratic Equations→Quadratic Equations — all question types→Maths STD 10 question papers→Gujarat Board STD 10 question papers→Gujarat Board question paper generator→

Similar questions

Comparing the equation $x(2 x-1)-5=0$ with $a x^2+b x+c=0, a=\ldots \ldots \ldots \quad(1,3,2)$

$\sqrt{5}$ is _______ number. (rational, irational, minus integer)

Normally, the condition that $a$ and $b$ both equally zero can be described by ............ .
$
\left(a^2-b^2 \neq 0, a^2+b^2 \neq 0,-a^2+b^2 \neq 0\right)
$

If roots of the equation $x^2-k x+4=0$ are equal then $k=\ldots \ldots .(4,-4, \pm 4)$

If the pair of linear equations $a_1 x+b_1 y+c_1=0$ and $a_2 x+ b _2 y+ c _2=0$ has unique solution then x=___________
$
\begin{array}{l}
{\left[\frac{ c _1, a _2- c _2 a _1}{ a _1, b _2- a _2 b _1}, \frac{ b _1 c _1- b _2 c _1}{ a _1 b _2- a _2 b _2}, \text { where } a_1 b_2-a_2 b_1 \neq 0 \text {, where } a, b_2-a_2 b_1 \neq 0 \frac{ a _1 b _2- a _2 b _1}{1}\right]}
\end{array}
$

In a box there are 6 Green, 3 white and 11 blue colour balls. One ball is selected at random. Then ………….. is the probability of an event that selected ball is NOT of blue colour. $\left(\frac{11}{20}, \frac{1}{20}, \frac{9}{20}\right)$

A quadrilateral ABCD circumscribes a circle. AB = 6 BC = 8 CD = 5 then AD =____________ .(3.11, 8)

If the graph of the linear equations $3 x+2 k y=2$ and $2 x+5 y+1=0$ represents parallel lines then $k=$ _____________$\left(\frac{15}{4}, \frac{4}{15}, \frac{3}{5}\right)$

$A (3,0), B (0,4)$ and $C (3,4)$ are vertices of $\triangle ABC$. $CD$ is a median. The coordinates of $D$ is _______ $\left(\left(2, \frac{3}{2}\right),\left(\frac{3}{2}, 2\right),\left(\frac{3}{2}, \frac{3}{2}\right)\right)$

If $\alpha$ and $\beta$ are zeroes at polynomial $x^2+x-\frac{1}{2}$ then $\frac{1}{\alpha}+\frac{1}{\beta}=\ldots \ldots \ldots . .(2,-2,1)$