Question
Find the value of $'a\ ':4x^2+ ax + 9 = (2x - 3)^2$
✓
Answer
$4x^2+ ax + 9 = (2x - 3)^2$
Comparing coefficients of $x$ terms, we get
$ax = -12x$
so$, a = -12$
Comparing coefficients of $x$ terms, we get
$ax = -12x$
so$, a = -12$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Draw a graph of each of the following equations$: x + 5 = 0$
→An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.
Factorise:
$\left(a^2-b^2\right)\left(c^2-d^2\right)-4 a b c d \quad\left[\right.$ Hint. Given Exp $\left.=(a c-b d)^2-(a d+b c)^2\right]$
→$\left(a^2-b^2\right)\left(c^2-d^2\right)-4 a b c d \quad\left[\right.$ Hint. Given Exp $\left.=(a c-b d)^2-(a d+b c)^2\right]$
Find the range of the data : 11, 13, 17, 14, 19, 14, 15, 18
Draw the graph of:
x + 4 = 0
x + 4 = 0
In the given figure, ABCD is a square, $EF \| BD$ and R is the mid-point of EF.
Prove that:
(i) $BE = DF$
(ii) AR bisects $\angle BAD$
(iii) If AR is produced, it will pass through C .
→Prove that:
(i) $BE = DF$
(ii) AR bisects $\angle BAD$
(iii) If AR is produced, it will pass through C .
Solve for $x : (a^{3x + 5})^2. (a^x)^4= a^{8x + 12}$
→Without using trigonometric tables, evaluate.
$\frac{\sin 80^{\circ}}{\cos 10^{\circ}}+\sin 59^{\circ} \sec 31^{\circ}$
→$\frac{\sin 80^{\circ}}{\cos 10^{\circ}}+\sin 59^{\circ} \sec 31^{\circ}$
The weights of 7 boys in a group are 52 kg, 57 kg, 55 kg, 60 kg, 54 kg, 59 kg and 55 kg. Find the mean weight of the group.
Without actual division, find which of the fractions are terminating decimal:
$\frac{23}{80}$
→$\frac{23}{80}$